<s>[INST] Partial derivatives problem, find $\frac{\partial z}{\partial x}$ Assume that $z$ and $w$ are differentiable functions of $x$ and $y$, satisfying the equations $xw^3+yz^2+z^3=-1$ and $zw^3-xz^3+y^2w=1.$ Then, at $\left(x,y,z,w\right)=\left(1,-1,-1,1\right)$, $\ \frac{\partial z}{\partial x}=$?
My solution 
$xw^3+yz^2+z^3=-1$
$\frac{\partial}{\partial x}\left(xw^3+yz^2+z^3\right)=0
\Rightarrow w^3+3xw^2\frac{\partial w}{\partial x}+2yz\frac{\partial z}{\partial x}+3z^2\frac{\partial z}{\partial x}=0$
$zw^3-xz^3+y^2w=1$ 
$\frac{\partial }{\partial x}\left(zw^3-xz^3+y^2w\right)=0
\Rightarrow w^3\frac{\partial z}{\partial x}+3zw^2\frac{\partial w}{\partial x}-z^3-3xz^2\frac{\partial z}{\partial x}+y^2\frac{\partial w}{\partial x}=0$
$\Rightarrow 
\begin{cases}
3xw^2\frac{\partial w}{\partial x}+\left(2yz+3z^2\right)\frac{\partial z}{\partial x}=-w^3 \\
\left(3zw^2+y^2\right)\frac{\partial w}{\partial x}+\left(w^3-3xz^2\right)\frac{\partial z}{\partial x}=z^3
\end{cases}$
When $\left(x,y,z,w\right)=\left(1,-1,-1,1\right)$, we have 
$ 
\begin{cases}
3\frac{\partial w}{\partial x}+5\frac{\partial z}{\partial x}=-1 \\
-2\frac{\partial w}{\partial x}-2\frac{\partial z}{\partial x}=-1 
\end{cases} 
$.
Therefore $\frac{\partial z}{\partial x}=-\frac{5}{4}$
Is there a problem with this solution? Or is this a better solution?
 [/INST] Bavo, your solution is good. The method is correct and simple.
If an even simpler method is wanted, Mr.Mathematician asks his friend Mr.Physicist. But the purists will not be fully happy :
$$\begin{cases}
xw^3+yz^2+z^3=-1\\
zw^3-xz^3+y^2w=1
\end{cases}\quad\to\quad
\begin{cases}
(1+dx)(1+dw)^3+(-1)(-1+dz)^2+(-1+dz)^3=-1\\
(-1+dz)(1+dw)^3-(-1+dx)(-1+dz)^3+(-1)^2(1+dw)=1
\end{cases}$$
After simplification with keeping only the first order variational terms :
\begin{cases}
-5dz-3dw=dx\\
2dz+2dw=dx
\end{cases}
Solving the linear  system for $dz$ and $dw$ :
$$\left(\begin{matrix}
dz\\
dw
\end{matrix}\right)=
\left(\begin{matrix}
-5 & -3\\
2 & 2
\end{matrix}\right)^{-1}
\left(\begin{matrix}
1\\
1
\end{matrix}\right)dx =
\left(\begin{matrix}
-\frac54\\
\frac74
\end{matrix}\right)dx $$
$$\quad\frac{\partial z}{\partial x}=-\frac54\quad \text{and} \quad\frac{\partial w}{\partial x}=\frac74$$
</s>