Answer
$$z=\frac{4}{5}+\frac{-12}{25}(x-\ln 4)+\frac{12}{25}(y-\ln 2)$$
Work Step by Step
Given $$f(x, y)=\operatorname{sech}(x-y), \quad(\ln 4, \ln 2)$$
Since
\begin{align*}
f(x,y)&= \operatorname{sech}(x-y) \ \ \ \ & f(\ln 4, \ln 2)&= \operatorname{sech}(\ln 2)=\frac{4}{5}\\
f_x(x,y)&=-\operatorname{sech}(x-y)\tanh (x-y)\ \ \ \ & f_x(\ln 4, \ln 2)&=\frac{-12}{25} \\
f_y(x,y)&= \operatorname{sech}(x-y)\tanh (x-y) \ \ \ \ & f_y(\ln 4, \ln 2)&= \frac{12}{25} \\ \\
\end{align*}
Then the tangent plane at $(\ln 4, \ln 2)$ is given by
\begin{align*}
z&= f(\ln 4, \ln 2)+f_{x}(\ln 4, \ln 2)(x-\ln 4)+f_{y}(\ln 4, \ln 2)(y-\ln 2)\\
&=\frac{4}{5}+\frac{-12}{25}(x-\ln 4)+\frac{12}{25}(y-\ln 2)
\end{align*}
