Answer
$$y =\frac{1}{2}(x-4)+\ln 16$$
Work Step by Step
Given $$f(x)=\ln \left(x^{2}\right) \quad x=4$$
Since at $x=4$, $f(x) = \ln 16$ and
$$f'(x)= \frac{2x}{x^2}= \frac{2}{x} $$
Then $ m= f'(x)\bigg|_{x=4}=\dfrac{1}{2} $
Hence, the tangent line is given by
\begin{align*}
\frac{y-y_1}{x- x_1}&=m\\
\frac{y-\ln 16 }{x- 4}&=\dfrac{1}{2} \\
y&=\frac{1}{2}(x-4)+\ln 16
\end{align*}
