Implicit differantiation Let z and w be differantiable functions of x and y and they satisfy the following equations.
$$ xw^3+yz^2+z^3=-1$$ $$ zw^3-xz^3+y^2w=1$$
Find $\frac{\partial z}{\partial x}$ and its value at $(x,y,z,w)=(1,-1,-1,1)$.
I don't know what to do with two implicit functions. So I thought maybe I can write $w$ in the first equation as a function of x,y and z such as :
$$ w = \left( -\frac{1+yz^2+z^3}{x} \right) ^{1/3}$$
Then I can put it into the second equation:
$$ z\left( \frac{-(1+yz^2+z^3)}{x} \right) - xz^3+y^2\left( \frac{-(1+yz^2+z^3)}{x} \right)^{1/3}=1$$
And here I can calculate the partial derivative as following:
$$F(x,y,z) = z\left( \frac{-(1+yz^2+z^3)}{x} \right) - xz^3+y^2\left( \frac{-(1+yz^2+z^3)}{x} \right)^{1/3} $$
$$\frac{\partial z}{\partial x} = - \frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}} $$
Then I found the result as:
$$ \frac{\partial z}{\partial x} =  \frac{\frac{z(1+yz^2+z^3)}{x^2}-z^3+\frac{y^2(1+yz^2+z^3)^{1/3}}{3x^{4/3}}}{\frac{-(4z^3+3yz^2+1)}{x}-3xz^2-\frac{y^2\left(\frac{z(3z+2y)}{3(z^3+yz^2+1)^{2/3}} \right)}{3x^{4/3}}}$$
But I'm guessing that I missed something here by eliminating $w$. Could you please give me your opinions? Thank you.
 A: Since you have
\begin{align}
&x w^3 + y z^2 + z^3 = -1 \\
&z w^3 - x z^3 + y^2 w = 1,
\end{align}
take the derivative with respect to $x$ for both equation, you have
\begin{align}
&x (3 w^2 \frac{\partial w}{\partial x}) + w^3 + y (2 z \frac{\partial z}{\partial x})
 + 3 z^2 \frac{\partial z}{\partial x} = 0 \\
&z (3 w^2 \frac{\partial w}{\partial x}) + (\frac{\partial z}{\partial x}) w^3 - 
 x (3 z^2 \frac{\partial z}{\partial x}) - z^3 + y^2 \frac{\partial w}{\partial x} = 0,
\end{align}
then by substituting $(x, y, z, w) = (1, -1, -1, 1)$, you have
\begin{align}
& 3 \frac{\partial w}{\partial x} + 1 + 2 \frac{\partial z}{\partial x}
 + 3 \frac{\partial z}{\partial x} = 0 \\
-&3 \frac{\partial w}{\partial x} + \frac{\partial z}{\partial x} - 
  3 \frac{\partial z}{\partial x} + 1 + \frac{\partial w}{\partial x} = 0.
\end{align}
Obviously, the solution for $\frac{\partial z}{\partial x}$ is
$$
\frac{\partial z}{\partial x}|_{(x,y,z,w)=(1,-1,-1,1)} = -\frac{5}{4}
$$
A: Indeed, you don't need to eliminate $w$.
Differentiating the first equation with respect to $x$ gives you
$$ w^3 + 3 xw^2 \frac{\partial w}{\partial x} + 2 y z\frac{\partial z}{\partial x} + 3 z^2 \frac{\partial z}{\partial x}=0$$  and with the replacement  $(x,y,z,w)=(1,-1,-1,1)$:
$$ 1 + 3  \frac{\partial w}{\partial x} + 5 \frac{\partial z}{\partial x}=0$$
Proceeding the same way by differentiating the second equation with respect to $x$, you get another linear equation linking $\frac{\partial z}{\partial x}$ and $\frac{\partial w}{\partial x}$.
You finally have a linear system of two equations linking $\frac{\partial z}{\partial x}$ and $\frac{\partial w}{\partial x}$. Solving it you get the desired $\frac{\partial z}{\partial x}$.
