Let $f$ be a differentiable function. Denote a bilinear form by $$b(f,f) = \int_{0}^{1} \bigg( \frac{d f(x)}{dx} \bigg)^{2} dx.$$ Given $f(0) = 0,$ we want to show that $$a \cdot b(f,f) \geq ||f||_{\infty}^{2},$$ where $||f||_{\infty} = \max_{0 \leq x \leq 1} |f(x)|.$ Note that $a$ is a constant independent of $f$.
In attempting the proof (which doesn't get far), I express $$||f||_{\infty}^{2} = \max_{0 \leq x \leq 1} |f(x)| \cdot |f(x)|.$$ However, I am not sure of a bound for this quantity that will lead naturally to the desired proof. Also, this might not be the first step in the correct proof.
Note: By Holder's inequality we have $$\bigg( \int_{0}^{1} \frac{d f (x)}{dx} dx \bigg)^{2} \leq \int_{0}^{1} \bigg( \frac{d f(x)}{dx} \bigg)^{2} dx.$$ The left hand side equals $(f(1) - f(0))^{2},$ which simplifies to $(f(1))^{2}$ since $f(0) = 0.$
Observe \begin{align} |f(x)| = \left|\int^x_0 f'(t)\ dt\right| \leq \int^1_0|f'(t)|\ dt \leq \left(\int^1_0|f'(t)|^2\ dt \right)^{1/2}. \end{align} Since this holds for all $x \in [0, 1]$ then you are done.