If $f$ is a bounded continues function, a function of $x^2$, $f^{\prime}(x)$ exists and $f$ is a pdf, so $f^{\prime}(0)=0$.prove or counter example

Question

If $f$ is a bounded continues function($\exists M\in R^+$ such that $\forall x\in R, \quad 0\leq f(x)<M$), $f$ is a function of $x^2$, $\forall x\in R, \quad f^{\prime}(x)$ exists and $f$ is a probability density function, so $$f^{\prime}(0)=0.$$

Prove or disprove by counterexample?

Is it valid to simple conclude $f^{\prime}(0)=0$, since $f^{\prime}(x)=2xg^{\prime}(x^2)$?