Let $f \colon \mathbb{R} \to \mathbb{R}$ be $\mathcal{C}^1[a,b]$ with bounded derivative. Let $H=L^2[a,b]$ and consider the Nemytskii operator $F\colon H \to H$ defined by $$F(x)(\psi)=f(x(\psi))$$ for $\psi \in [a,b]$. Then I need to show that the Gateaux derivative $DF \colon H \to \mathcal{L}(H)$ of F is $$DF(x) h =\frac{\partial f}{\partial x} (x) h$$ for $h \in H$ being the direction. With the last inequality i mean that $$DF(x) h (\psi) =\frac{\partial f}{\partial x} (x(\psi)) h(\psi)$$ for every $\psi \in [a,b]$ so that $DF(x)h$ is the Nemystkii operator associated with the derivative of $f$.
I take $\psi \in [a,b]$ fixed and I compute $$\lim_{\alpha \to 0}\frac{F(x+\alpha h) - F(x)}{\alpha} (\psi)=\lim_{\alpha \to 0}\frac{f(x(\psi)+\alpha h(\psi)) - f(x(\psi))}{\alpha}= \frac{\partial f}{\partial x} (x(\psi))h(\psi)$$ for every $\psi$.
Then can I conclude that if I define a Nemitskii operator $DF(x)h$ by $DF(x) h (\psi)=\frac{\partial f}{\partial x} (x(\psi))$ this is the Gateaux derivative of $F$ in direction $h$?
I mean by the previous equality the following $$\lim_{\alpha \to 0}\frac{F(x+\alpha h) - F(x)}{\alpha} $$ is a Nemistkii operator and its action on $\psi$ is the same as $DF(x)h$ so that $$\lim_{\alpha \to 0}\frac{F(x+\alpha h) - F(x)}{\alpha} =DF(x)h$$ and then $DF(x)h$ is the Gateaux derivative.
Let $x,h\in L^2$. We need to prove that $$ \left\|\frac1t(f(x+th)-f(x)-h f'(x)h)\right\|_{L^2(a,b)}^2= \int_a^b \left(\frac{f(x(s)+th(s))-f(x(s))-tf'(x(s))h(s)}t\right)^2 ds \to 0 $$ for $t\searrow0$. Since $f$ is $C^1$, the integrand tends pointwise to zero for almost all $s$. Since $f'$ is a bounded function, $|f'(y)|\le M$ for all $y$, we have by the mean-value theorem the integrable upper bound of the integrand: $$ \left(\frac{f(x(s)+th(s))-f(x(s))-tf'(x(s))h(s)}t\right)^2 \le \left(\frac{Mth(s)+Mth(s)}t\right)^2 \le 4 M^2 h(s)^2. $$ Now the claimed convergence follows from dominated convergence theorem. This proves the formula for the directional derivative. It is now easy to check that $$ h\mapsto f'(x(\cdot))h $$ is a linear and continuous mapping, which proves Gateaux differentiability.