I have rewritten this answer. We actually need a stronger version of Hartshorne's Theorem 4.6A (the theorem of the primitive element).
Theorem 4.6A$^\star$. Let $L$ be a finite separable extension field of a field $K$, and suppose that $K$ contains an infinite subset $S$. Then, there is an element $\alpha \in L$ which generates $L$ as an extension field of $K$. Furthermore, if $\beta_1,\beta_2,\ldots,\beta_n$ is any set of generators of $L$ over $K$, then $\alpha$ can be taken to be a linnear combination
$$\alpha = c_1\beta_1 + c_2\beta_2 + \cdots + c_n\beta_n$$
of the $\beta_i$ with coefficients $c_i \in S$.
Proof. This follows from the proof of [Zariski–Samuel, Ch. II, §9, Thm. 19], which uses Kronecker's "method of indeterminates," but we rewrite their proof below.
Consider the field extension
$$L \subseteq L(X,X_1,X_2,\ldots,X_n)$$
of $L$, where $X,X_1,X_2,\ldots,X_n$ are a set of indeterminates, and consider the subfields
\begin{align*}
K^\star &= K(X_1,X_2,\ldots,X_n)\\
L^\star &= L(X_1,X_2,\ldots,X_n)
\end{align*}
in $L(X,X_1,X_2,\ldots,X_n)$. Then, $L^\star = K^\star(\beta_1,\beta_1,\ldots,\beta_n)$, and $L^\star$ is a finite separable extension of $K^\star$ since the $\beta_i$ are separable over $K$, and hence also separable over $K^\star$ (see [Zariski–Samuel, Ch. II, §5, Lem. 2]). Consider the element
$$\beta^\star = X_1\beta_1 + X_2\beta_2 + \cdots + X_n\beta_n \in L^\star.\tag{1}\label{eq:zs1}$$
Let $F(X)$ be the minimal polynomial of $\beta^\star$ in $K^\star[X]$. The coefficients of $F(X)$ are rational functions of $X_1,X_2,\ldots,X_n$ with coefficients in $K$; let $g(X_1,X_2,\ldots,X_n) \in K[X_1,X_2,\ldots,X_n]$ be a common denominator of these rational functions. Then,
$$g(X_1,X_2,\ldots,X_n) \cdot F(X) = f(X,X_1,X_2,\ldots,X_n) \in K[X,X_1,X_2,\ldots,X_n],$$
and we have
$$f(\beta^\star,X_1,X_2,\ldots,X_n) = 0.\tag{2}\label{eq:zs2}$$
Let
$$G(X_1,X_2,\ldots,X_n) = f(X_1\beta_1+X_2\beta_2+\cdots+X_n\beta_n,X_1,X_2,\ldots,X_n).\tag{3}\label{eq:zs3}$$
Then, $G(X_1,X_2,\ldots,X_n)$ is a polynomial in $X_1,X_2,\ldots,X_n$ with coefficients in $L$, and \eqref{eq:zs2} says $G(X_1,X_2,\ldots,X_n) = 0$. Thus, all partial derivatives $\partial G/\partial X_i$ are zero for $i \in \{1,2,\ldots,n\}$. By \eqref{eq:zs3}, we then have
$$\beta_i \cdot f'(\beta^\star,X_1,X_2,\ldots,X_n) + f_i(\beta^*,X_1,X_2,\ldots,X_n) = 0\tag{4}\label{eq:zs4}$$
for every $i \in \{1,2,\ldots,n\}$, where
\begin{align*}
f'(X,X_1,X_2,\ldots,X_n) &= \frac{\partial f(X,X_1,X_2,\ldots,X_n)}{\partial X},\\
f_i(X,X_1,X_2,\ldots,X_n) &= \frac{\partial f(X,X_1,X_2,\ldots,X_n)}{\partial X_i}.
\end{align*}
The left-hand side in each equation \eqref{eq:zs4} is a polynomial in $L[X_1,X_2,\ldots,X_n]$ by \eqref{eq:zs1}, and hence is the zero polynomial. Thus, the equations \eqref{eq:zs4} remain valid if we substitute for $X_1,X_2,\ldots,X_n$ any elements of $K$. On the other hand, we have
$$f'(X,X_1,X_2,\ldots,X_n) = g(X_1,X_2,\ldots,X_n)\,F'(X)$$
where $F'(X) = dF/dX$, and hence
$$f'(\beta^\star,X_1,X_2,\ldots,X_n) \ne 0,$$
since $\beta^\star$ is separable over $K^\star$ and therefore $F'(\beta^\star) \ne 0$. Thus, $f'(\beta^\star,X_1,X_2,\ldots,X_n)$ is a nonzero polynomial in $L[X_1,X_2,\ldots,X_n]$. Since $S \subseteq L$ and $S$ is an infinite subset, we can find elements $c_1,c_2,\ldots,c_n \in S$ such that $(c_1,c_2,\ldots,c_n)$ is not a zero of that polynomial [Zariski–Samuel, Ch. I, §18, Thm. 14]. Setting
$$\beta = c_1\beta_1 + c_2\beta_2 + \cdots + c_n\beta_n,$$
we have that
$$f'(\beta,c_1,c_2,\ldots,c_n) \ne 0\tag{5}\label{eq:zs5}$$
and
$$\beta_i \, f'(\beta,c_1,c_2,\ldots,c_n) + f_i(\beta,c_1,c_2,\ldots,c_n) = 0\tag{6}\label{eq:zs6}$$
for every $i \in \{1,2,\ldots,n\}$. The equation \eqref{eq:zs6} and the inequality \eqref{eq:zs5} imply that $\beta_i \in K(\beta)$, and since $\beta \in L$, we therefore see that $L = K(\beta)$. This completes the proof of the theorem. $\blacksquare$
We now prove Hartshorne's exercise, which we restate below.
Exercise [Hartshorne, Ch. I, Exer. 4.9]. Let $X$ be a projective variety of dimension $r$ in $\mathbf{P}^n$, with $n\geq r+2$. Show that for a suitable choice of $P \notin X$, and a linear $\mathbf{P}^{n-1} \subseteq \mathbf{P}^n$, the projection from $P$ to $\mathbf{P}^{n-1}$ induces a birational morphism of $X$ onto its image $X' \subseteq \mathbf{P}^{n-1}$.
Proof. Let $k$ denote the ground field over which $X$ is defined. After permuting coordinates, we may assume without loss of generality that $X\cap U_0\neq\emptyset$. Then, the images of the rational functions $x_i/x_0$ generate $K(X)$ over $k$. Since $k$ is algebraically closed, we see the extension $K(X)/k$ is separably generated by [Hartshorne, Ch. I, Thm. 4.8A]. Since $\dim X = r$, after possible permutation of coordinates, we have that $x_1/x_0,x_2/x_0,\ldots,x_r/x_0$ form a separating transcendence basis for $K(X)$ over $k$ by [Hartshorne, Ch. I, Thm. 4.7A] to give the chain
$$k \subseteq k(x_1/x_0,x_2/x_0,\ldots,x_r/x_0) \subseteq K(X)$$
of field extensions.
Next, by setting $S = k$ in the theorem of the primitive element (Theorem 4.6A$^\star$ above), we see that $K(X)$ is generated by
$$\alpha = \sum_{i=r+1}^n c_i\frac{x_i}{x_0}$$
where $c_i \in k$ for every $i$. After a linear change of coordinates, we may assume that $\alpha = x_{r+1}/x_0$.
Now consider the map
\begin{align*}
\mathbf{P}^n &\dashrightarrow \mathbf{P}^n\\
[x_0:\cdots:x_{n-1}:x_n] &\mapsto [x_0:\cdots:x_{n-1}:0]
\end{align*}
This is the projection away from the point $P = Z(x_0,x_1,\ldots,x_{n-1})$ to the hyperplane $Z(x_n)$. Restricting the codomain to $Z(x_n) \simeq \mathbf{P}^{n-1}$, we obtain the rational map
\begin{align*}
\pi\colon \mathbf{P}^n &\dashrightarrow \mathbf{P}^{n-1}\\
[x_0:\cdots:x_{n-1}:x_n] &\mapsto [x_0:\cdots:x_{n-1}]
\end{align*}
which we claim induces a birational map $\pi\rvert_X\colon X \dashrightarrow \pi(X)$.
Let $X'$ denote the image of $X$ through this map $\pi$. Then, $K(X')$ is generated by the rational functions $x_i/x_0$ for $1 \le i \le r+1$. The map on function fields corresponding to $\pi$ is
\begin{align*}
K(X') &\hookrightarrow K(X)\\
x_i/x_0 &\mapsto x_i/x_0
\end{align*}
Since $x_i/x_0$ for $1 \le i \le r+1$ generate $K(X)$, we see that $K(X') = K(X)$, and so the map $\pi$ is birational. $\blacksquare$
We note that to prove Exercise 3.14 with this method, we can perform the operation above $n - (r+1)$ times to find a birational map between $X$ and a hypersurface in $\mathbf{P}^{r+1}$. In the notation of the proof above, composing all of these projections from points can be described as the linear projection
\begin{align*}
\mathbf{P}^n &\dashrightarrow \mathbf{P}^n\\
[x_0:\cdots:x_n] &\mapsto [x_0:\cdots:x_{r+1}:0:0\cdots:0]
\end{align*}
away from the $(n-r-1)$-plane $Z(x_1,x_2,\ldots,x_{r+1})$ to the $(r+1)$-plane $Z(x_{r+2},x_{r+3},\ldots,x_n)$; see for example [Shafarevich, Ex. 1.27]. Restricting the codomain to the $(r+1)$-plane $Z(x_{r+2},x_{r+3},\ldots,x_n) \simeq \mathbf{P}^{r+1}$, this gives a rational map $\pi\rvert_X\colon X \dashrightarrow \mathbf{P}^{r+1}$ that is birational onto its image.
