\[
\begin{align}
z &\depends (x, y)
\\
z &= x^2 - y^2
\end{align}
\]
Let's compute the metric differential.
\[
\begin{align}
\hat{\diff} z &= \hat{\diff} \bigl(x^2 - y^2\bigr)
\\
&= \hat{\diff} \bigl(x^2\bigr) - \hat{\diff} \bigl(y^2\bigr)
\\
&= 2 x \mult \hat{\standard}_{x} - 2 y \mult \hat{\standard}_{y}
\end{align}
\]
We've found the tangent metric.
\[
\begin{align}
\tangent z &= 2 x \mult \hat{\standard}_{x} - 2 y \mult \hat{\standard}_{y}
\end{align}
\]
Let's see what the tangent metric looks like at the point \( (3, 1) : \Point_{(x, y)} \).
\[
\begin{align}
\tangent z &= 6 \mult \hat{\standard}_{x} - 2 \mult \hat{\standard}_{y}
\end{align}
\]
At \( (3, 1) \), the tangent plane has a positive \( x \)-slope and a negative \( y \)-slope. Can you find a point where both the \( x \)-slope and the \( y \)-slope are positive? Or a point where the tangent plane is horizontal?