Answer
$f(g(x))=x-\frac{7}{10}$
$g(f(x))=x-\frac{7}{6}$
$f^{-1}(x)=\frac{5}{3}x-\frac{5}{6}$ $\ne$ $g(x)$
Therefore, $g(x)$ is not an Inverse of $f(x)$.
Work Step by Step
Given $$f(x)=\frac{3}{5}x+\frac{1}{2}$$ $$g(x)=\frac{5}{3}x-2.$$
Calculate $f(g(x))$ and $g(f(x))$:
$$f(g(x))=\frac{3}{5}(\frac{5}{3}x-2)+\frac{1}{2}=x-\frac{6}{5}+\frac{1}{2}=x-\frac{7}{10}$$ $$g(f(x))=\frac{5}{3}(\frac{3}{5}x+\frac{1}{2})-2=x+\frac{5}{6}-2=x-\frac{7}{6}$$
As $$f(g(x))\not=x,g(f(x))\not=x$$
it follows that $g(x)$ is not an Inverse of $f(x)$.
