The chain rule for functionals

Question

I want to show the following equality. \begin{align*} \partial G(J,K)(\eta,y)=\frac{\partial G}{\partial J}(J(y),K(y))\delta J(\eta,y)+\frac{\partial G}{\partial K}(J(y),K(y))\delta K(\eta,y) \end{align*} Here, $J(y)$ and $K(y)$ functionals defined by \begin{align*} J(y)=\int_{x_0}^{x_1}f(x,y,y')dx \end{align*} and \begin{align*} K(y)=\int_{x_0}^{x_1}g(x,y,y')dx, \end{align*} and $\delta$ is defined by \begin{align*} \delta J(y,\eta)=\int_{x_0}^{x_1}\left(\frac{\partial f}{\partial y}\eta+\frac{\partial f}{\partial y'}\eta'\right)dx, \end{align*} where $y\in S$ and $\eta\in H$ for $$S=\{y\in C^2[x_0,x_1]:y(x_0)=y_0\wedge y(x_1)=y_1\}$$ and $$H=\{\eta\in C^2[x_0,x_1]:\eta(x_0)=\eta(x_1)=0\}.$$ My attempts this far have been $$\delta G(\eta,y)=\int_{x_0}^{x_1}\left(\left(\frac{\partial f}{\partial y}\circ g\right)\frac{\partial g}{\partial y}\eta+\left(\frac{\partial f}{\partial y'}\circ g\right)\frac{\partial g}{\partial y'}\eta'\right)dx,$$ but I don't really know how to procede from here. I don't really know what $$\frac{\partial G}{\partial J}(J(y),K(y))\delta J(\eta,y)+\frac{\partial G}{\partial K}(J(y),K(y))\delta K(\eta,y)$$ looks like written out in terms of these functionals, so I don't even know exactly what I want to obtain. Any help with this is greatly appreciated.