Solving $u(f,g)/v(f,g)=f'$, where $f,g \in k[t]$

Question

Let $k$ be a field of characteristic zero and let $f=f(t),g=g(t) \in k[t]$. Denote $f'=f'(t)$ the derivative of $f$. Let $u=u(x,y),v=v(x,y) \in k[x,y]$, with $v \neq 0$, and $u,v$ coprime in $k[x,y]$. (I do not mind to further assume that $k \in \{\mathbb{R},\mathbb{C}\}$, if this simplifies the answer).

Is it possible to find a general solution to $u(f,g)/v(f,g)=f'$?

Although $u(x,y)$ and $v(x,y)$ are assumed to be coprime in $k[x,y]$, it may happen (or it must happen?) that $u(f(t),g(t))$ and $v(f(t),g(t))$ are not coprime in $k[t]$; for example, $u(x,y)=2y, v(x,y)=x, f(t)=t^2, g(t)=t^3$.

Ways for solution: (1) Integrate both sides, but I do not know what the result on the left side should be. (2) Differentiate both sides: Then I guess we should apply the chain rule, but still I am not sure I know how to solve the new equation. (3) Perhaps some considerations of degrees are relevant here (perhaps no). (4) I thought that this question is relevant, but now I see that it may not be relevant, since here $u(f(t),g(t))$ and $v(f(t),g(t))$ are probably not coprime in $k[t]$.

Example: $f(t)=t^6+t^2$, $g(t)=t^3$. We have, $\frac {g}{f-g^2}= \frac{t^3}{t^6+t^2-t^6}= \frac{t^3}{t^2}=t$. Therefore, $g'(t)=3t^2=3(\frac {g}{f-g^2})^2$ and $f'(t)=6t^5+2t=6(\frac {g}{f-g^2})^5+2(\frac {g}{f-g^2})$. I do not understand how the below answer helps in such cases.

Thank you very much for any help!

Answer 1

Hint: $g$ is completely unconstrained. What happens to the degrees when we take $g = 0$? or $g = f$?

Further hint: $g = 0$: $\deg_x u(x,y) \deg_t f = (\deg_t(f) - 1) + \deg_x v(x,y) \deg_t f$. Look at this equation modulo $\deg_t f \geq 2$ to determine that $\deg_t f$ is very constrained. So a general solution is out. I might also mention that the cases where a solution is possible are constrained away as well by conditions on $u$ and $v$.