I wish to prove that $$\mathbb{C}[T] \cong \frac{\mathbb{C} [X,Y, T]}{ (Y^2 - X^2(X+1), XT-Y, T^2 - X)}. $$
How does one in general prove things like this? I would say that $$ \frac{\mathbb{C} [X,Y, T]}{ (Y^2 - X^2(X+1), XT-Y, T^2 - X)} \cong \frac{ \mathbb{C}[X,T]}{( T^2 - X)} \cong \mathbb{C}[T] $$ but this looks more like guess work to me than an actual proof.
Since $\mathbb{C}[X,Y,T]$ is a polynomial algebra, any algebra morphism from $\mathbb{C}[X,Y,T]$ to a commutative algebra is uniquely determined by the images of the three generators. Thus, to find a candidate isomorphism $\mathbb{C}[X,Y,T]/(Y^2-X^2(X+1),XT-Y,T^2-X)\to\mathbb{C}[Z]$ you need to find three elements in $\mathbb{C}[Z]$ that satisfy the given relations.
Of course, by looking at the last two relations, $$T\mapsto P(Z), \\ X\mapsto P(Z)^2, \\ Y\mapsto P(Z)^3.$$ When do these satisfy the third relation? Let's check $$Y^2-X^2(X+1) \mapsto P(Z)^6-P(Z)^4(P(Z)^2+1) = -P(Z)^4.$$ The only option would be $P(Z)=0$, which of course does not give an isomorphism. Conclusion, no isomorphism between the two.
Abstractly speaking, there is no way to have an isomorphism between them: $\mathbb{C}[Z]$ is a domain, while on the other side we have that $t^6 = y^2 = t^6+t^4$ implies $0 = t^4 = t^2\cdot t^2$ and $t^2=x\neq 0$.