If $u^3+v+w=x+y^2+z^2,u+v^3+w=x^2+y+z^2,u+v+w^3=x^2+y^2+z,$ prove the following I am stuck with the following problem that says : 

If$$\begin{align}
u^3+v+w&=x+y^2+z^2,\\u+v^3+w&=x^2+y+z^2,\\u+v+w^3&=x^2+y^2+z,
\end{align}$$then prove that 
  $$\frac{\partial(u,v,w)}{\partial(x,y,z)}=\frac{1-4(xy+yz+zx)+16xyz}{2-3(u^2+v^2+w^2)+27u^2v^2w^2}$$

My try:  I have to compute the value of $$\begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} &\frac{\partial u}{\partial z} \\ 
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} &\frac{\partial v}{\partial z} \\ 
\frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} &\frac{\partial w}{\partial z}
\end{vmatrix}$$.
For that I need to convert $u,v,w$ in terms of $x,y,z$. I  am unable to do that as things are getting complicated. 
From the given set of equations we get, $$3u^2 \frac{\partial u}{\partial x}+ \frac{\partial v}{\partial x}+ \frac{\partial w}{\partial x}=1 \\
 \frac{\partial u}{\partial x}+ 3v^2\frac{\partial v}{\partial x}+ \frac{\partial w}{\partial x}=2x \\
 \frac{\partial u}{\partial x}+ \frac{\partial v}{\partial x}+ 3w^2\frac{\partial w}{\partial x}=2x \implies\\ 
\begin{pmatrix}
3u^2 & 1 &1\\ 
 1&  3v^2& 1\\ 
 1& 1 & 3w^2
\end{pmatrix} \begin{pmatrix}
\frac{\partial u}{\partial x}\\ 
\frac{\partial v}{\partial x}\\ 
\frac{\partial w}{\partial x}
\end{pmatrix}=\begin{pmatrix}
1\\ 
2x\\ 
2x
\end{pmatrix}$$
Can someone  give me a detailed explanation? Thanks in advance for your time.

You've already expressed the partial derivatives with respect to $x$ as a matrix-vector equation:
$$\begin{pmatrix}
3u^2 & 1 &1\\ 
 1&  3v^2& 1\\ 
 1& 1 & 3w^2
\end{pmatrix} \begin{pmatrix}
\frac{\partial u}{\partial x}\\ 
\frac{\partial v}{\partial x}\\ 
\frac{\partial w}{\partial x}
\end{pmatrix}=\begin{pmatrix}
1\\ 
2x\\ 
2x
\end{pmatrix}.$$
Doing the same for partial derivatives with respect to $y$ and $z$, we get
$$\begin{pmatrix}
3u^2 & 1 &1\\ 
 1&  3v^2& 1\\ 
 1& 1 & 3w^2
\end{pmatrix} \begin{pmatrix}
\frac{\partial u}{\partial y}\\ 
\frac{\partial v}{\partial y}\\ 
\frac{\partial w}{\partial y}
\end{pmatrix}=\begin{pmatrix}
2y\\ 
1\\ 
2y
\end{pmatrix}$$
and
$$\begin{pmatrix}
3u^2 & 1 &1\\ 
 1&  3v^2& 1\\ 
 1& 1 & 3w^2
\end{pmatrix} \begin{pmatrix}
\frac{\partial u}{\partial z}\\ 
\frac{\partial v}{\partial z}\\ 
\frac{\partial w}{\partial z}
\end{pmatrix}=\begin{pmatrix}
2z\\ 
2z\\ 
1
\end{pmatrix}.$$
The key step remaining is to combine these three into a single matrix-matrix equation:
$$\begin{pmatrix}
3u^2 & 1 &1\\ 
 1&  3v^2& 1\\ 
 1& 1 & 3w^2
\end{pmatrix} \begin{pmatrix}
\frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}&\frac{\partial u}{\partial z}\\ 
\frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}&\frac{\partial v}{\partial z}\\ 
\frac{\partial w}{\partial x}&\frac{\partial w}{\partial y}&\frac{\partial w}{\partial z}
\end{pmatrix}=\begin{pmatrix}
1&2y&2z\\ 
2x&1&2z\\ 
2x&2y&1
\end{pmatrix}.$$
Taking determinants, we get
$$(2-3(u^2+v^2+w^2)+27u^2v^2w^2)\left\vert\frac{\partial(u,v,w)}{\partial(x,y,z)}\right\vert=1-4(xy+yz+zx)+16xyz$$
and the result follows immediately.
