Partial derivatives

It is given that:

$\begin{cases} w = x^2+y^2+z^2 \\ z^3-xy+yz+ y^3 = 1 \end{cases} \Rightarrow \begin{cases} \dfrac {\partial w} {\partial x} = 2x+2z \dfrac {\partial z} {\partial x} & ...(1) \\ 3z^2\dfrac {\partial z} {\partial x} -y+y\dfrac {\partial z} {\partial x} = 0 & ... (2) \end{cases}$

Substituting $x=2$ , $y=-1$ and $z=1$ in Eq. 2:

$\begin{aligned} 3(1)\frac {\partial z} {\partial x} -(-1)+(-1)\frac {\partial z} {\partial x} & = 0 \\ 2\frac {\partial z} {\partial x} + 1 & = 0 \\ \Rightarrow \frac {\partial z} {\partial x} & = -\frac{1}{2} \end{aligned}$

From Eq. 1: $\begin{aligned} \frac {\partial w} {\partial x} & = 2(2)+2(1) \left( -\frac {1} {2} \right) = \boxed{3} \end{aligned}$

$\begin{cases} w = x^2+y^2+z^2 \\ z^3-xy+yz+ y^3 = 1 \end{cases} \Rightarrow \begin{cases} \dfrac {\partial w} {\partial x} = 2x+2z \dfrac {\partial z} {\partial x} & ...(1) \\ 3z^2\dfrac {\partial z} {\partial x} -y+y\dfrac {\partial z} {\partial x} = 0 & ... (2)\\\therefore~ \dfrac {\partial z} {\partial x} =\dfrac y { 3z^2+y} & ...(3) \end{cases} \\Eliminating ~~\dfrac{\partial z}{\partial x} ~\text{from (1) and (3), and substituting for (x,y,z),}\\\dfrac {\partial w} {\partial x}=2x+2z*\dfrac y { 3z^2+y} =2*2+2*1*\dfrac {-1}{3*1^2-1} =~~~~ \large \color{#D61F06}{3}$