$\newcommand{\d}{\mathrm{d}}$I am having a problem here because acceleration varies with distance, and I am not able to relate it with time
\begin{align} a &=-\frac{kx}{m} \\ v\frac{\d v}{\d x} &= -\frac{kx}{m} \end{align} For $x_{\max}$ \begin{align} 0-v_{0}^2 &= -\frac{kx^2}{m} \\ x &= \sqrt{\frac{mv_0}{k}} \end{align}
But I still can’t find time
$\newcommand{\d}{\mathrm{d}}$We have, $$ v \,\d v =-Cx \,\d x \\ \int_{v_0}^{v(x)} v \,\d v = -C\int_0^x x\,\d x \\ \frac{v(x)^2-v_0^2}{2}=-Cx \\ v(x)=\sqrt{v_0^2-2Cx}, $$ where $C=\frac km$. Using $v=\frac{\d x}{\d t}$, $$\frac{\d x}{\sqrt{v_0^2-2Cx}} = \d t$$ Now just integrate both sides to get an expression for $x(t)$, after which you need to solve $x(t)=0$.