A question regarding the chain rule for partial derivatives Let $f: \mathbb{R^2} \to \mathbb {R}$ be a differentiable function and consider the function $F:\mathbb{R^3}\to \mathbb{R}, F(x, y, z)=f(x^2-y+2yz^2, z^3e^{xy})$. Compute $\frac{\partial F}{\partial x}$, $\frac{\partial F}{\partial y}$ and $\frac{\partial F}{\partial z}$ in terms of $f$'s first order partial derivatives.
I began by recognising that $F=f\circ g$, where $g:\mathbb{R}^3 \to \mathbb{R^2}, g(x, y, z)=(x^2-y+2yz^2, z^3e^{xy})$. Let's denote by $u(x,y,z):=x^2-y+2yz^2$ and $v(x,y,z)=z^3e^{xy}$ $g$'s components.
By the chain rule I know that $$\frac{\partial F}{\partial x}(x,y,z)=\frac{\partial f}{\partial u}(x^2-y+2yz^2, z^3e^{xy})\cdot \frac{\partial u}{\partial x}(x,y,z)+ \frac{\partial f}{\partial v}(x^2-y+2yz^2, z^3e^{xy})\frac{\partial v}{\partial x}(x,y,z)$$ and the same relations hold for $\partial y$ and $\partial z$, but I don't understand how/if I could further simplify $\frac{\partial f}{\partial u}(x^2-y+2yz^2, z^3e^{xy})$ and $\frac{\partial f}{\partial v}(x^2-y+2yz^2, z^3e^{xy})$. As far as I understand, these are the partials derivatives of $f$ with respect to the functions $u$ and $v$. How do I compute these?
 A: To make things clearer, denote $u$ and $v$ the variables for $f$, where
$$u=x^2-y+2yz^2,\qquad v=z^3\mathrm e^{xy}.$$
The chain rule asserts that
\begin{align}
\frac{\partial F(x,y,z)}{\partial x}&=\frac{\partial f(u,v )}{\partial u}\biggl|_{\substack{u=x^2-y+2yz^2\\v=z^3\mathrm e^{xy}}}\!\cdot \frac{\partial u(x,y,z)}{\partial x}+\frac{\partial f(u,v)}{\partial v}\biggl|_{\substack{u=x^2-y+2yz^2\\v=z^3\mathrm e^{xy}}}\!\cdot\frac{\partial v(x,y,z)}{\partial x} \\
&=\frac{\partial f(u,v)}{\partial u}\biggr|_{\substack{u=x^2-y+2yz^2\\v=z^3\mathrm e^{xy}}}\!\cdot 2x+\frac{\partial f(u,v)}{\partial v}\biggr|_{\substack{u=x^2-y+2yz^2\\v=z^3\mathrm e^{xy}}}\!\cdot yz^3\mathrm e^{xy}
\end{align}
and similarly for the other partial derivatives.
A: If you use the chain rule for the derivative of $multivariate$ functions, you can read off the $partial$ derivatives. More precisely, following your idea, we have
$F'(x_0,y_0,z_0)=(f\circ g)'(x_0,y_0,z_0)=f'(g(x_0,y_0,z_0))\circ g'(x_0,y_0,z_0).$
In matrix form,
$\begin{pmatrix}
F_x(x_0,y_0,z_0) &F_y(x_0,y_0,z_0)  &F_z (x_0,y_0,z_0)
\end{pmatrix}=$
$\begin{pmatrix}
f_x(g(x_0,y_0,z) & f_y(g(x_0,y_0,z_0)
\end{pmatrix}\begin{pmatrix}
2x_0 &2z_0^2-1  & 2y_0z_0\\ 
 y_0z_0^3e^{x_0y_0}& x_0z_0^3e^{x_0y_0} & 3z_0^2e^{x_0y_0}
\end{pmatrix}$
Now multilpy the matrices to read off the derivatives.
