Answer
\[
2=y^{\prime}(0)
\]
Work Step by Step
We find:
$\frac{\partial z}{\partial x}(1,2)=4, \frac{\partial z}{\partial y}(1,2)=2$
Also, $x=x(t), y=y(t)$ are differentiable functions of $t$ with
$1=x(0), 2=y(0), x^{\prime}(0)=-\frac{1}{2}\ \text{and} \ 2=z^{\prime}(0)$
\[
\begin{array}{c}
z^{\prime}(t)=\frac{d z}{d t}=\frac{\partial z}{\partial x} \cdot \frac{d x}{d t}+\frac{\partial z}{\partial y} \cdot \frac{d y}{d t} \\
z^{\prime}(t)=\frac{d z}{d t}=\frac{\partial z}{\partial x} \cdot x^{\prime}(t)+\frac{\partial z}{\partial y} \cdot y^{\prime}(t) \\
z^{\prime}(0)=\frac{\partial z}{\partial x}(1,2) \cdot x^{\prime}(0)+\frac{\partial z}{\partial y}(1,2) \cdot y^{\prime}(0) \\
(4) \cdot-\frac{1}{2}+2 \cdot y^{\prime}(0) =2\\
\therefore\left[2=y^{\prime}(0)\right]
\end{array}
\]
