I will write a very detailed answer. Suppouse our ellipsoid is given by $f^{-1}(1)$ where $f(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}$. Any curve passing through points in the preimage of $1$ are such that $f\circ \alpha(s)=1$. Differentiating with respect to $s$ yields $\nabla f(\alpha(s))\cdot \alpha'(s)=0$. As $\nabla f(\alpha(s))\not=0$ we may take normalize it to gain a Gauss Normal Map $N=\frac{(x/a^2,y/b^2,z/c^2)}{\sqrt{x^2/a^4+y^2/b^4+z^2/c^4}
}$ where $h=\sqrt{x^2/a^4+y^2/b^4+z^2/c^4}$.
$$h(\alpha(s))N(\alpha(s))={\left(\frac{\alpha_1}{a^2},\frac{\alpha_2}{b^2},\frac{\alpha_3}{c^2}\right)}{}\Rightarrow $$
$$(h\circ\alpha)'(s)N(\alpha(s))+h\circ\alpha(s) DN(\alpha'(s))=\left(\frac{\alpha_1'}{a^2},\frac{\alpha_2'}{b^2},\frac{\alpha_3'}{c^2}\right) \tag{1}$$
Let $\alpha'=(v_1,v_2,v_3)$.
Claim: A point is umbilic if and only if $DN(v)\cdot(N\times v)=0$
Proof: $(\Rightarrow)$ Being umbilic, $DN(v)=\lambda v$ and the results holds.
$(\Leftarrow)$ $\{v, N, N\times v\}$ forms an orthogonal basis if $v\in T_pS$ is not equal to zero. $DN(v)=\lambda_v v+\lambda_2 N+\lambda_3 N\times v$. Furthermore $\lambda_2=0$ ( because $DN(v)\in T_pS$) and $\lambda_3=0$ ( because $DN(v)\cdot(N\times v)=0$). Thus $DN(v)=\lambda_v v$. In particular if $e_1$ and $e_2$ are principal directions:
$$\lambda_{e_1+e_2}(e_1+e_2)=DN(e_1+e_2)=DN(e_1)+DN(e_2)=k_1e_1+k_2e_2$$
Thus, as $\{e_1,e_2\}$ is a basis $k_1=k_2=\lambda_{e_1+e_2}$.
With this lemma combined with equation (1), we must have umbilic points if and only if $DN(v)\cdot(v\times N)=0$, which happens if and only if:
$$\left(\frac{\alpha_1'}{a^2},\frac{\alpha_2'}{b^2},\frac{\alpha_3'}{c^2}\right) \cdot(v\times N)=0 \:\text{ where $v=\alpha'\in T_pS$} \Leftrightarrow $$
$$ \frac{1}{h}\text{det}\begin{bmatrix}\frac{v_1}{a^2} & \frac{v_2}{b^2} & \frac{v_3}{c^2}\\
\frac{x}{a^2} & \frac{y}{b^2} & \frac{z}{c^2}\\
v_1 & v_2 & v_3 \end{bmatrix}=0 \:\text{ where $v\in T_p S$ and $(x,y,z)=p$} \Leftrightarrow$$
$$\frac{-x}{a^2}\left(\frac{v_2v_3}{b^2}-\frac{v_2v_3}{c^2}\right) +\frac{y}{b^2}\left(\frac{v_1v_3}{a^2}-\frac{v_1v_3}{c^2}\right)-\frac{z}{c^2}\left(\frac{v_1v_2}{a^2}-\frac{v_1v_2}{b^2}\right)=0\tag{I}$$
$$ v_1 \frac{x}{a^2}+v_2 \frac{y}{b^2}+v_3 \frac{z}{c^2}=0\tag{II}$$
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 \tag{III}$$
Thus, umbilical points in an ellipsoid must satisfy these three equations. Let us suppose $0<a<b<c$.
Claim: $z=0$ yields no solutions.
Proof: by equation (III), $z=0$ implies that either $x\not=0$ or $y\not=0$. Let us take $x\not=0$. By equation (II) $v_2$ and $v_3$ may assume any value but $v_1$ is determined if $v\in T_pS$. Furthermore, by multiplying equation (I) by $x$ and taking $v_3,v_2\not =0$:
$$ \frac{x^2}{a^2}v_2v_3\left(\frac{1}{b^2}-\frac{1}{c^2}\right)=\frac{yx}{b^2}v_1v_3\left(\frac{1}{a^2}-\frac{1}{c^2}\right)=\frac{-v_1}{v_2}\frac{x^2}{a^2}v_1v_3\left(\frac{1}{a^2}-\frac{1}{c^2}\right)\Rightarrow $$
$$\frac{x^2}{a^2}v_2^2\left(\frac{1}{b^2}-\frac{1}{c^2}\right)=-v_1^2\frac{x^2}{a^2}\left(\frac{1}{a^2}-\frac{1}{c^2}\right)$$
This equation implies that a number strictly larger than $0$ is also non positive. Contradiction! Similarly, if we had supposed $y\not=0$, by equation (II) $v_1$ and $v_3$ may assume any value but $v_2$ is determined if $v\in T_pS$. Furthermore, by multiplying equation (I) by $y$ and taking $v_1,v_2\not =0$:
$$\frac{y^2}{b^2}v_1v_3\left(\frac{1}{a^2}-\frac{1}{c^2}\right)= \frac{xy}{a^2}v_2v_3\left(\frac{1}{b^2}-\frac{1}{c^2}\right)=\frac{-v_2}{v_1}\frac{y^2}{b^2}v_2v_3\left(\frac{1}{b^2}-\frac{1}{c^2}\right)\Rightarrow $$
$$\frac{y^2}{a^2}v_1^2\left(\frac{1}{a^2}-\frac{1}{c^2}\right)=-v_2^2\frac{y^2}{b^2}\left(\frac{1}{b^2}-\frac{1}{c^2}\right)$$
Thus, a strictly greater than $0$ number is also non positive, which is absurd.
We notice somethings. Firstly, $v_1$ and $v_2$ may be taken to be any real number, as long as we fix $v_3$ to be that for which equation (II) is satisfied. Furthermore, With $z$ being non-negative, equation (I) will hold if and only if it also holds when multiplied by $\frac{z}{c^2}$. This yields:
$$\frac{-xz}{a^2c^2}\left(\frac{v_2v_3}{b^2}-\frac{v_2v_3}{c^2}\right) +\frac{yz}{b^2c^2}\left(\frac{v_1v_3}{a^2}-\frac{v_1v_3}{c^2}\right)-\frac{z^2}{c^4}\left(\frac{v_1v_2}{a^2}-\frac{v_1v_2}{b^2}\right)=0 $$
By using equation (II) this further yields that equations (I) and (II) will hold if and only if:
$$v_2^2\frac{yx}{b^2a^2}\left(\frac{1}{b^2}-\frac{1}{c^2}\right)-v_1^2\frac{yx}{b^2a^2}\left(\frac{1}{a^2}-\frac{1}{c^2}\right)+v_1v_2\left(\frac{x^2}{a^4}\left(\frac{1}{b^2}-\frac{1}{c^2}-\right)-\frac{y^2}{b^4}\left(\frac{1}{a^2}-\frac{1}{c^2}-\right)-\frac{z^2}{c^4}\left(\frac{1}{a^2}-\frac{1}{b^2}\right)\right) =0$$
If $v_1=0$ and $v_2=1$, we must have $xy=0$. If $x=0$, we have a positive number $\frac{z^2}{c^4}\left(\frac{1}{a^2}-\frac{1}{b^2}\right)$ being non-positive $-\frac{y^2}{b^4}\left(\frac{1}{a^2}\right)$ which is absurd. Thus $y=0$ and we must have (I) (II) and (III) holding if and only if $y=0$ and:
$$\frac{x^2}{a^2}(c^2-b^2)=\frac{z^2}{c^2}(b^2-a^2)\quad \text{and} \quad \frac{x^2}{a^2}+\frac{z^2}{c^2}=1\Leftrightarrow$$
$$\begin{cases}
x^2=a^2\frac{b^2-a^2}{c^2-a^2}\\
y=0\\
z^2=c^2\frac{c^2-b^2}{c^2-a^2}
\end{cases}$$
