Answer
At point $P$:
$\frac{{\partial f}}{{\partial x}}{|_P}$ is positive.
$\frac{{\partial f}}{{\partial y}}{|_P}$ is negative.
Work Step by Step
We have $f\left( {x,y} \right) = z$.
Referring to Figure 7, we consider the plane that is parallel to the $xz$-plane. Notice that the vertical trace that passes through the point $P$ has positive slope in $xz$-coordinates at $P$, therefore the partial derivatives $\frac{{\partial f}}{{\partial x}}{|_P}$ is positive.
Using Figure 7 we consider the vertical trace that passes through the point $P$, which lies in the plane parallel to $yz$-plane. The slope at $P$ on this trace curve is negative in $yz$-coordinates, therefore the partial derivatives $\frac{{\partial f}}{{\partial y}}{|_P}$ is negative.
