Answer
$$z=\frac{8}{3} x-\frac{2}{3} y+(\ln 3-2) $$
Work Step by Step
Given $$f(x, y)=\ln \left(4 x^{2}-y^{2}\right), \quad(1,1)$$
Since
\begin{align*}
f(x,y)&=\ln \left(4 x^{2}-y^{2}\right) \ \ \ \ & f(1,1)&=\ln 3 \\
f_x(x,y)&=\frac{8x}{4 x^{2}-y^{2}} \ \ \ \ & f_x(1,1)&= \frac{8}{3} \\
f_y(x,y)&= \frac{-2y}{4 x^{2}-y^{2}} \ \ \ \ & f_y(1,1)&= \frac{-2}{3}\\
\end{align*}
Then the tangent plane at $(1,1)$ is given by
\begin{align*}
z&= f(1,1)+f_{x}(1,1)(x-1)+f_{y}(1,1)(y-1)\\
&= \ln 3+\frac{8}{3}(x-1)+\frac{-2}{3}(y-1)\\
&=\frac{8}{3} x-\frac{2}{3} y+(\ln 3-2)
\end{align*}
