Answer
$$
P_y= \frac{y}{\sqrt{ y^2+z^2}}e^\sqrt{ y^2+z^2},\\
P_z = \frac{z}{\sqrt{ y^2+z^2}}e^\sqrt{ y^2+z^2}.
$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall that $(x^n)'=nx^{n-1}$
Since $ P=e^\sqrt{ y^2+z^2}$, then by using the chain rule, we have
$$
P_y=e^\sqrt{ y^2+z^2} \frac{2y}{2\sqrt{ y^2+z^2}}= \frac{y}{\sqrt{ y^2+z^2}}e^\sqrt{ y^2+z^2},\\
P_z=e^\sqrt{ y^2+z^2} \frac{2z}{2\sqrt{ y^2+z^2}}= \frac{z}{\sqrt{ y^2+z^2}}e^\sqrt{ y^2+z^2}.
$$
