Counterexample: a smoothed version of the function $f$ given by the formula $$f(x)=(h/2)^{1/2}(1-\max(0,|x/h-1|))\tag{1}\label{1}$$ with $h\downarrow0$.
Indeed, if we had $$\|f\|_\infty\le C\,\|f\|_{2}^{2/3} \tag{2}\label{2}$$ for some real $C>0$ and all continuously differentiable functions $f\colon[0,1]\to\Bbb R$ such that $f(0)=f(1)=0$ and $\|f'\|_2\le1$, then, by approximation, we would have \eqref{2} for $f$ as in \eqref{1}, because for such $f$ we have $f(0)=f(1)=0$ and $\|f'\|_2=1$. However, for such $f$, we also have $\|f\|_\infty=(h/2)^{1/2}$, whereas $\|f\|_{2}^{2/3}\asymp h^{2/3}=o(h^{1/2})$ as $h\downarrow0$.
This shows that the best (largest) exponent $p>0$ for which $$\|f\|_\infty\le C\,\|f\|_{2}^p \tag{2a}\label{2a}$$ can hold (for some real $C>0$ and all continuously differentiable functions $f\colon[0,1]\to\Bbb R$ such that $f(0)=f(1)=0$ and $\|f'\|_2\le1$) is $\le1/2$.
On the other hand, for any continuously differentiable function $f\colon[0,1]\to\Bbb R$ such that $f(0)=f(1)=0$ and $\|f'\|_2\le1$ and any $x\in[0,1]$, $$f(x)^2=\int_0^x dt\,\frac{d}{dt}\,f(t)^2 =2\int_0^x dt\,f(t)f'(t)\le2\|f\|_2\|f'\|_2 \le2\|f\|_2,$$ so that \eqref{2a} holds with $p=1/2$ and $C=\sqrt2$. Thus, the best $p$ in \eqref{2a} is $1/2$.