If $f(tx,ty,tz) = t^nf(x,y,z)$ then in my lecture notes it says differentiating this equation with respect to $t$ gives:
$$x\frac{df}{dx} + y\frac{df}{dy} + z\frac{df}{dz} = nt^{n-1}f(x,y,z)$$
But why isn't it:
$$x\frac{df}{d(tx)} + y\frac{df}{d(ty)} + z\frac{df}{d(tz)} = nt^{n-1}f(x,y,z)\ ?$$
Suppose $f(tx,ty,tz)=t^nf(x,y,z)$. Consider the chain rule: $$ \frac{d}{dt} f(x_1(t),x_2(t),x_3(t)) = \frac{\partial f}{\partial x_1}(x_1(t),x_2(t),x_3(t))\frac{dx_1}{dt}+\frac{\partial f}{\partial x_2}(x_1(t),x_2(t),x_3(t))\frac{dx_2}{dt}+\frac{\partial f}{\partial x_3}(x_1(t),x_2(t),x_3(t))\frac{dx_3}{dt} $$ I am being really pedantic here, but the question raised involves this issue. The partial derivatives are properly understood to be evaluated at the inside function which happens to be $(x_1(t),x_2(t),x_3(t))$ in this context. Returning to the given identity, identify $x_1=tx$ and $x_2=ty$ and $x_3=tz$ where presumably $x,y,z$ are not functions of time (if they were we'd obtain a different identity). Note: $$ \frac{dx_1}{dt} = x, \ \ \frac{dx_2}{dt} = y, \ \ , \frac{dx_3}{dt} = z $$ Consequently, $$ \frac{d}{dt} f(x_1(t),x_2(t),x_3(t)) = x\frac{\partial f}{\partial x_1}(x_1(t),x_2(t),x_3(t))+y\frac{\partial f}{\partial x_2}(x_1(t),x_2(t),x_3(t))+z\frac{\partial f}{\partial x_3}(x_1(t),x_2(t),x_3(t)) $$ which means, $$ \frac{d}{dt} f(tx,ty,tz) = x\frac{\partial f}{\partial x_1}(tx,ty,tz)+y\frac{\partial f}{\partial x_2}(tx,ty,tz)+z\frac{\partial f}{\partial x_3}(tx,ty,tz) $$ So, I tend to think your book takes some liberty with notation.
Your thoughts are mostly correct, but just now written clearly.
Normally, a function in three variables is denoted $f(x, y, z)$, and the partial derivatives are $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$, $\frac{\partial f}{\partial z}$, what they mean is the partial derivative with respect to each variable.
To interpret $f(tx, ty, tz)$ nicely, we shall write $f$ as a function $f(x_1, x_2, x_3)$, and $x_1(t) = tx, x_2(t) = ty, x_3(t) = tz$, so chain rule gives
$$ \frac{d}{dt}f(x_1(t), x_2(t), x_3(t)) = \frac{dx_1}{dt}\frac{\partial f}{\partial x_1} + \frac{dx_2}{dt}\frac{\partial f}{\partial x_2} + \frac{dx_3}{dt}\frac{\partial f}{\partial x_3}. $$
So you are confused with $x, y, z$ denoting the first, second, third terms of $f$, and the fact that are used as other variables again.
The notes is alright but not clear enough, you are in the right direction as well, but your notation is not the best way to illustrate it.
The $\partial f/\partial x$ in your notes is rather taking derivative with respect to the first variable than taking derivative with respect to $x$.
Try to think in this way. Say you have a function with three slots to fill (notice this function satisfy the relation you gave, so it may serve as an example): $$ f(\square, \triangle,\bigcirc) = \square^n + \triangle^n + \bigcirc^n.\tag{1} $$ Now we fill the first variable by $x(t)$, the other two untouched then: $$ \frac{df}{dt} = \frac{\partial }{\partial \square}f(\square, \triangle,\bigcirc) \Bigg\vert_{\square = x(t)} \cdot \frac{d x(t)}{dt}, $$ the first term is taking derivative with respect to the first variable evaluated at $x(t)$. Taking (1) as an example: let $x(t) = tx$ $$ \frac{df}{dt} = \frac{\partial }{\partial \square}f(\square, \triangle,\bigcirc) \Bigg\vert_{\square = tx} \cdot x = n \square^{n-1}\big|_{\square = tx} \cdot x = nt^{n-1}x^n, $$ If we evaluate directly $$ f(tx, \triangle,\bigcirc) = t^n x^n + \triangle^n + \bigcirc^n, $$ you will get the same result as chain rule.
Keeping this in mind your formula can be written in a least ambiguous way: for $f(u,v,w)$
$$x\frac{\partial f}{\partial u}\Bigg\vert_{u = tx} + y\frac{\partial f}{\partial v} \Bigg\vert_{v = ty} + z\frac{\partial f}{\partial w}\Bigg\vert_{w = tz} = nt^{n-1}f(x,y,z).$$
