Solving this ODE through derivatives?

I'm currently trying to solve a more complex system of ODE. In order to understand the environment more, I've simplified stuff a bit, such that I have arrived at

$$\alpha F(x) = G(x) + F_x(x)\cdot(H(x, F(x))$$

where I need to solve for $F$. $G$ and $H$ are known, and $\alpha > 0$. I can I can write this equation as - with $A_i$ denoting different non-zero constants -

$$\alpha F(x) = \frac{A_0}{(A_0 + xA_1)^2}A_5 + F_x(x) \cdot \left[(1-x)A_2 + x \left(1-\left(\frac{A_4}{F(x)}\right)^\phi\right)^\frac{1}{\phi}\right]$$

My plan was to take the derivative w.r.t. $x$ several times, and hope that something cancels out, such that I get a clear representation for the i-th derivative of $F_x$, and then would work backwards.

However, even the central term, doesn't cancel out eventually as far as I can see.

$$G(x) = \frac{A_0}{(A_0 + xA_1)^2}\\ G_x(x) = \frac{-2A_0 (A_0 + xA_1)A_1}{(A_0 + xA_1)^4} = \frac{-2A_0A_1}{(A_0 + xA_1)^3} = -2A_1\frac{G(x)}{(A_0 + xA_1)}\\ G_{xx}(x) = \frac{-2A_0A_1 3 (A_0 + xA_1)^2A_1}{(A_0 + xA_1)^6} = \frac{-2\cdot3A_0A_1A_1}{(A_0 + xA_1)^4} = 3A_1\frac{G_x(x)}{(A_0 + xA_1)}\\ G_{xxx}(x) = \frac{-2A_0A_1A_1 3 \cdot 4(A_0 + xA_1)^3A_1}{(A_0 + xA_1)^{8}} = \frac{-2A_0A_1A_1A_1 4\cdot 3}{(A_0 + xA_1)^5} = \frac{4A_1G_{xx}(x)}{(A_0 + xA_1)}$$

Such that, in general, for the ith partial derivative ($i > 1$), I believe it is true that

$$\frac{\partial^i G(x)}{\partial x^i} = \frac{(i+1)A_1}{(A_0 + xA_1)} \frac{\partial^{i-1} G(x)}{\partial x^{i-1}}$$

This means that I can never take sufficiently many derivatives of $F(x)$ w.r.t. $x$ and hope that eventually stuff cancels out, or does it? How should I approach this problem?