If $\int f(x) \sin x \cos x d x=\frac{1}{2\left(a^2-b^2\right)} \log |f(x)|+C$, then $f(x)=$

$\frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x}$

Let

$I =\int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} \sin x \cos x d x$

$\Rightarrow I =\frac{1}{2\left(a^2-b^2\right)} \int \frac{\left(a^2-b^2\right) \sin 2 x}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x$

$\Rightarrow I =\frac{1}{2\left(a^2-b^2\right)} \int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d\left(a^2 \sin ^2 x+b^2 \cos ^2 x\right)$

$\Rightarrow I=\frac{1}{2\left(a^2-b^2\right)} \log \left|a^2 \sin ^2 x+b^2 \cos ^2 x\right|+C$

Hence, $f(x)=\frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x}$