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?

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 $dw$ :
$$\left(\begin{matrix}
dz\\
dw
\end{matrix}\right)=
\left(\begin{matrix}
-5 & -3\\
2 & 2
\end{matrix}\right)^{-1}
\left(\begin{matrix}
1\\
1
\end{matrix}\right)dx =
\left(\begin{matrix}
-\frac54\\
\frac74
\end{matrix}\right)dx $$
$$\quad\frac{\partial z}{\partial x}=-\frac54\quad \text{and} \quad\frac{\partial w}{\partial x}=\frac74$$
