Answer
\[
t+1=x, 3=y, 6 t+3=z
\]
Work Step by Step
We are given that
\[
x^{2} y=z(x, y)
\]
$y=z$
\[
\begin{array}{l}
\frac{\partial}{\partial y} z(x, y)=z_{y}(x, y) \\
\frac{\partial}{\partial y}(y)=z_{y}(x, y) \\
1=z_{y}(x, y)
\end{array}
\]
$j+k$ is parallel to the tangent line at (1,3,3) Thus, the parametric equation is
\[
t+3=x, t+3=y,1=x
\]
b) Finding the intersection point of the curve $z(x, y)$ and $3=y$, we get $3 x^{2}=z$
\[
\begin{aligned}
&\frac{\partial}{\partial y} z(x, y)= z_{x}(x, y) \\
&\frac{\partial}{\partial y}\left(3 x^{2}\right)=z_{x}(x, y) \\
6 x=z_{x}(x, y) & \\
\left.\frac{\partial z}{\partial y}\right|_{(x=1)}=6 &
\end{aligned}
\]
$i+6 k$ is parallel to the tangent line at (1,3,3); thus, the parametric equation is
\[
t+1=x, 3=y, 6 t+3=z
\]
