Answer
$$9y+6\sqrt{e}z-5x=35$$
Work Step by Step
Given $$x^{2}+z^{2} e^{y-x}=13, \quad P=\left(2,3, \frac{3}{\sqrt{e}}\right)$$
Consider $f(x,y,z)=x^{2}+z^{2} e^{y-x}-13 $, since
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle\\
&=\left\langle 2x-z^2e^{y-x},z^2e^{y-x},2ze^{y-x} \right\rangle\\
\nabla f_{P}&=\langle -5, 9, 6\sqrt{e}\rangle
\end{align*}
Then the equation of the tangent plane at $P$ is given by
\begin{align*}
\nabla f_{P} \cdot\langle x-x_1, y-y_1, z-z_1\rangle&= 0\\
\langle -5, 9, 6\sqrt{e}\rangle\cdot\langle x-2, y-3, z-\frac{3}{\sqrt{e}}\rangle&=0\\
-5(x-2)+9(y-3)+ 6\sqrt{e}(z-\frac{3}{\sqrt{e}})&=0\\
9y+6\sqrt{e}z-5x&=35
\end{align*}
