$\mathbb C [x,y] \to \mathbb C [t]$ with the evaluation : $$ \text{ev}_{x = t^2, y = t^3} $$
How can you show that the kernel is principal in $\mathbb C[x,y] $? I think the kernel is the polynomial ideal :
$$x^3 - y^2$$ which is irreducible (Irreducibility of polynomial $x^3-y^2$)
now, how can I prove that $$\ker( \text{ev} ) \subset (x^3 - y^2)$$
I was thinking about an euclidean division, but I'm stuck.
It is indeed clear that $x^3-y^2\in\ker(\mathrm{ev})$ and so $(x^3-y^2)\subset\ker(\mathrm{ev})$. You already note that $x^3-y^2$ is irreducible, so if $\ker(\mathrm{ev})$ is principal then it follows that $\ker(\mathrm{ev})=(x^3-y^2)$.
Let $f\in\ker(\mathrm{ev})$ so that $f(t^2,t^3)=0$. Then the remainder of $f$ when divided by $x^3-y^2$ (as a polynomial in $y$ with coefficients in $\Bbb{C}[x]$) is of the form $$r=r_1(x)y+r_0(x),$$ for some $r_0r_1\in\Bbb{C}[x]$. Then we have $$0=r(t^2,t^3)=r_1(t^2)t^3+r_0(t^2).$$ All terms of $r_0(t^2)$ are of even degree, and all terms of $r_1(t^2)t^3$ are of odd degree, so for the above to equal $0$ we must have $r_0=r_1=0$. This shows that $r=0$ and hence $f\in(x^3-y^2)$, and so $\ker\mathrm{ev}=(x^3-y^2)$.