Setup:
Suppose that $q(x)$ is a strictly positive $C^{2}$ function and consider the initial value problem: \begin{equation} \begin{aligned} y''+q(x)y=&0\\ y(0)=&k\\ y'(0)=&c \end{aligned} \end{equation} We know that such an initial value problem has a unique solution which we will denote by $f^q(x)$.
Define the functional $F_{x}$ by
\begin{equation} \begin{aligned} F_{x}:C^{2}(\mathbb{R};\mathbb{R})&\rightarrow \mathbb{R}\\ %C^{1,2}(\mathbb{R}^2;\mathbb{R})\\ F_{x}(q)&\mapsto f^{q}(x). \end{aligned} \end{equation}
Question:
Is the functional $F_{x}$ Frechet differentiable? How can we determine if it is?
Fix $q_{0}\in C^{2}$. To find the Gateaux derivative let $p\in C^{2}$ and consider $q_{0}+tp$ where $t\in\mathbb{R}$. Consider \begin{gather*} y^{\prime\prime}+(q_{0}(x)+tp(x))y(x)=0,\\ y(0)=k,\quad y^{\prime}(0)=c. \end{gather*} The $y$ depends on $x$, $t$, $q_{0}$, $p$, so we can write it as $y=y(x,t;q_{0},p)$. In turn, we have \begin{gather*} \frac{\partial^{2}y}{\partial x^{2}}(x,t;q_{0},p)+(q_{0}(x)+tp(x))y(x,t;q_{0}% ,p)=0,\\ y(0,t;q_{0},p)=k,\quad\frac{\partial y}{\partial x}(0,t;q_{0},p)(0)=c. \end{gather*} Formally, if we take the derivative with respect to $t$ at $t=0$ we get \begin{gather*} \frac{\partial^{3}y}{\partial x^{2}\partial t}(x,0;q_{0},p)+(q_{0}% (x)+0p(x))\frac{\partial y}{\partial t}(x,0;q_{0},p)+1p(x)y(x,0;q_{0},p)=0,\\ \frac{\partial y}{\partial t}(0,0;q_{0},p)=0,\quad\frac{\partial^{2}% y}{\partial x\partial t}(0,0;q_{0},p)=0. \end{gather*} So if we write $y_{q_{0}}(x):=y(x,0;q_{0},p)$ and $z_{p}(x):=\frac{\partial y}{\partial t}(x,0;q_{0},p)$, we have that $z$ solves the differential equation \begin{gather*} z_{p}^{\prime\prime}+q_{0}z_{p}+py_{q_{0}}=0,\\ z_{p}(0)=0,\quad z_{p}^{\prime}(0)=0. \end{gather*} This can actually made rigorous.
So if we consider $F_{x}:C^{2}\rightarrow C$ and fix $q_{0}$ and a direction $p$, the conjecture is that the directional derivative $\frac{\partial F_{x}% }{\partial p}(q_{0})=z_{p}(x)$. If this is the case, then the Frechet derivative at $q_{0}$ is given by $dF_{x}(q_{0})(p)=\frac{\partial F_{x}% }{\partial p}(q_{0})=z_{p}(x)$ and so we need to prove that$$ \lim_{q\rightarrow q_{0}}\frac{F_{x}(q)-F_{x}(q_{0})-dF_{x}(q_{0})(q-q_{0}% )}{\Vert q-q_{0}\Vert_{C^{2}}}=0. $$ Let $y_{q}$ be the solution of the differential equation \begin{gather*} y_{q}^{\prime\prime}+q(x)y_{q}(x)=0,\\ y_{q}(0)=k,\quad y_{q}^{\prime}(0)=c. \end{gather*} Then we can rewrite the limit as$$ \lim_{q\rightarrow q_{0}}\frac{y_{q}(x)-y_{q_{0}}(x)-z_{q-q_{0}}(x)}{\Vert q-q_{0}\Vert_{C^{2}}}=0, $$ where $z_{q-q_{0}}$ solves \begin{gather*} z_{q-q_{0}}^{\prime\prime}+q_{0}z_{q-q_{0}}+(q-q_{0})y_{q_{0}}=0,\\ z_{q-q_{0}}(0)=0,\quad z_{q-q_{0}}^{\prime}(0)=0. \end{gather*} Since $y_{q}^{\prime\prime}+qy_{q}=0$, $y_{q_{0}}^{\prime\prime}+q_{0}% y_{q_{0}}=0$, and $z_{q-q_{0}}^{\prime\prime}+q_{0}z_{q-q_{0}}+(q-q_{0}% )y_{q_{0}}=0$, by subtracting the differential equations we get \begin{align*} y_{q}^{\prime\prime}-y_{q_{0}}^{\prime\prime}-z_{q-q_{0}}^{\prime\prime} & =-qy_{q}+q_{0}y_{q_{0}}+q_{0}z_{q-q_{0}}+(q-q_{0})y_{q_{0}}\\ & =-q(y_{q}-y_{q_{0}}-z_{q-q_{0}})+(q_{0}-q)z_{q-q_{0}}% \end{align*} with $(y_{q}-y_{q_{0}}-z_{q-q_{0}})(0)=0$ and $(y_{q}-y_{q_{0}}-z_{q-q_{0}% })^{\prime}(0)=0$. By integration$$ (y_{q}-y_{q_{0}}-z_{q-q_{0}})(x)=\int_{0}^{x}\int_{0}^{s}-q(t)(y_{q}-y_{q_{0}% }-z_{q-q_{0}})(t)+(q_{0}-q)(t)z_{q-q_{0}}(t)\,dtds $$ and so$$ |(y_{q}-y_{q_{0}}-z_{q-q_{0}})(x)|\leq\int_{0}^{x}\int_{0}^{s}|q(t)||(y_{q}% -y_{q_{0}}-z_{q-q_{0}})(t)|+|(q_{0}-q)(t)z_{q-q_{0}}(t)|\,dtds. $$ It follows by Gronwall's inequality that% \begin{align*} |(y_{q}-y_{q_{0}}-z_{q-q_{0}})(x)| & \leq\int_{0}^{x}\int_{0}^{s}% |(q_{0}-q)(t)z_{q-q_{0}}(t)|\,dtds\exp\left( \int_{0}^{x}\int_{0}% ^{s}|q(t)||(y_{q}-y_{q_{0}}-z_{q-q_{0}})(t)|\,dtds\right) \\ & \leq\Vert q-q_{0}\Vert_{C^{2}}\int_{0}^{x}\int_{0}^{s}|z_{q-q_{0}% }(t)|\,dtds\exp\left( \int_{0}^{x}\int_{0}^{s}|q(t)||(y_{q}-y_{q_{0}% }-z_{q-q_{0}})(t)|\,dtds\right) \end{align*} and so$$ \frac{|(y_{q}-y_{q_{0}}-z_{q-q_{0}})(x)|}{\Vert q-q_{0}\Vert_{C^{2}}}\leq \int_{0}^{x}\int_{0}^{s}|z_{q-q_{0}}(t)|\,dtds\exp\left( \int_{0}^{x}\int% _{0}^{s}|q(t)||(y_{q}-y_{q_{0}}-z_{q-q_{0}})(t)|\,dtds\right) . $$ It remains to prove that the right hand-side goes to zero as $q\rightarrow q_{0}$. By integration$$ z_{q-q_{0}}(x)=\int_{0}^{x}\int_{0}^{s}-q_{0}(t)z_{q-q_{0}}(t)+(q_{0}% -q)(t)y_{q_{0}}(t)\,dtds $$ and so by Gronwall's inequality \begin{align*} |z_{q-q_{0}}(x)| & \leq\int_{0}^{x}\int_{0}^{s}|(q_{0}-q)(t)y_{q_{0}% }(t)|\,dtds\exp\left( \int_{0}^{x}\int_{0}^{s}|q_{0}(t)||\,dtds\right) \\ & \leq\Vert q-q_{0}\Vert_{C^{2}}L. \end{align*} Hence, $\Vert z_{q-q_{0}}\Vert_{\infty}\leq\Vert q-q_{0}\Vert_{C^{2}% }L\rightarrow0$ as $q\rightarrow q_{0}$. I am skipping the proof that $y_{q}$ is bounded.