Let $u,v,d \in \mathbb{C}[t]$ be three polynomials such that $\deg(u) \geq 1,\deg(v) \geq 1,\deg(d) \geq 2$.
Assume that $\gcd(u,v)=\gcd(u,d)=\gcd(v,d)=1$, and denote $U=du$ and $V=dv$.
Further assume that there exists $H(x,y) \in \mathbb{C}[x,y]$ with $H(0,0)=0$ (= without constant term) satisfying $H(U,V)=0$.
Is it true that necessarily $H=0$?
A special case: If $\deg(H)=1$, then $H=\lambda x+\mu y$, hence $0=H(U,V)=\lambda du+ \mu dv= d(\lambda u + \mu v)$, so $\lambda u + \mu v =0$ which is impossible, unless $\lambda=\mu=0$ (otherwise, $\lambda u= - \mu v$ which implies that $\gcd(u,v)=u$, contradicting $\gcd(u,v)=1$).
However, if $\deg(H) \geq 2$, things may be more complicated.
There is a nice result which says the following: Given nonconstant $f,g \in \mathbb{C}[t]$: $f,g$ are algebraically dependent over $\mathbb{C}$ if and only if $f,g \in \mathbb{C}[h]$, for some $h \in \mathbb{C}[t]$; see this or Corollary 1.3 (for one direction).
By the nice result we know that $U,V \in \mathbb{C}[h]$, for some $h \in \mathbb{C}[t]$. Please notice that I assume a specific algebraic dependence, namely, one which satisfies $H(0,0)=0$; hopefully, this fact is relevant.
Thank you very much!
Let $d=t^2$, $u=t+1$, and $v=t-1$. Then $U+V=2t^3$ and $U-V=2t^2$, so $U$ and $V$ satisfy $H(U,V)=0$ where $H(x,y)=2(x+y)^2-(x-y)^3$.
In fact, such a nonzero $H$ always exists. Indeed, since $U$ and $V$ are algebraically dependent there exists $H\neq 0$ such that $H(U,V)=0$. Now just observe that the constant term of such an $H$ must be $0$, since otherwise $H(U,V)$ would be nonzero mod $d$.