Answer
$$z=x e^{2}-2 y e^{2}+e^{2}$$
Work Step by Step
Given $$g(x, y)=e^{x / y}, \quad(2,1)$$
Since
\begin{align*}
g(x, y)&= e^{x / y} \ \ \ \ & g(2,1)&=e^2 \\
g_x(x,y)&= \frac{1}{y}e^{x / y} \ \ \ \ & g_x(2,1)&= e^2\\
g_y(x,y)&=\frac{-x}{y^2}e^{x / y} \ \ \ \ & g_y(2,1)&= -2e^2\\
\end{align*}
Then the tangent plane at $(2,1)$ is given by
\begin{aligned}
z &=g(2,1)+g_{x}(2,1)(x-2)+g_{y}(2,1)(y-1) \\
&=e^{2}+e^{2}(x-2)-2 e^{2}(y-1)\\
&=x e^{2}-2 y e^{2}+e^{2}
\end{aligned}
