The homomorphism theorem says that from a surjective homomorphism $\phi\colon R\to S$ we can define an isomorphism
$$
\tilde{\phi}\colon R/\ker\phi\to S
$$
by mapping each coset $r+\ker\phi$ to $\phi(r)$.
If $x,y\in \ker\phi$, you have to show that
$$
(1+x)(1+y)\in 1+\ker\phi
$$
and that every element in $1+\ker\phi$ has an inverse which still belongs to $1+\ker\phi$. This ensures $1+\ker\phi$ is a subgroup of $R^*$.
For normality, you have to show that, for $x\in\ker\phi$ and $r\in R^*$, $r(1+x)r^{-1}\in 1+\ker\phi$.
Next, you have to show that $\phi^*\colon R^*\to S^*$ (the restriction of $\phi$) induces a morphism $R^*/N\to S^*$, that is, $N\subseteq \ker\phi^*$.
Just for completeness, here's a sketch.
If $x,y\in\ker\phi$, then $(1+x)(1+y)=1+(x+y+xy)\in 1+\ker\phi=N$, because $x+y+xy\in\ker\phi$.
If $x\in\ker\phi$, set $y=-x$; then we know that $y^n=0$, for some $n>0$. Then
$$(1-y)(1+y+y^2+\dots+y^{n-1})=1-y^n=1$$
and so $1-y=1+x$ is invertible (the element is obviously also a left inverse).
If $x\in\ker\phi$ and $r\in R^*$, then $r(1+x)r^{-1}=1+rxr^{-1}\in N$.
Now $\phi$ induces a group morphism $\phi^*\colon R^*\to S^*$, because invertibles in $R$ are mapped to invertible elements of $S$ (because $\phi$ is surjective). It is obvious that $N\subset\ker\phi^*$, because $\phi^*(1+x)=\phi(1+x)=1+\phi(x)=1$.
Conversely, if $r\in\ker\phi^*$, then $\phi(r)=1$, so $r-1\in\ker\phi$ and $r=1+(r-1)\in N$.
Observe that in general, if $d = \gcd(a,b)$, then $x^{b/d} - y^{a/d} \in \ker(\phi)$, so $\langle x^{b/d} - y^{a/d} \rangle \subseteq \ker(\phi)$.
To show the reverse inclusion, notice that by treating $\mathbb{F}[x,y]$ as $\mathbb{F}[y][x]$, using the division algorithm we can write any $f \in \mathbb{F}[x,y]$ as $q(x,y) (x^{b/d} - y^{a/d}) + r(x,y)$ where each term in the remainder $r(x,y)$ has $x$-degree less than $b/d$. Now, $\phi(f) = \phi(r)$. On the other hand, each possible monomial $x^m y^n$ with $0 \le m < b/d$ maps to a distinct monomial $z^{am + bn}$ (see below for a proof), so if $f \in \ker(\phi)$ then $\phi(r) = 0$ implies $r = 0$.
So, all we have left to show is the easy purely number theoretic result:
The map $\mathbb{N}_0 \times \mathbb{N}_0 \to \mathbb{N}_0, (m, n) \mapsto am + bn$, is injective when restricted to $[0, b/d) \times \mathbb{N}_0$.
To show this, suppose $am + bn = am' + bn'$; then $(a/d) (m-m') = (b/d) (n'-n)$. Since $\gcd(a/d, b/d) = 1$, this implies that $m \equiv m' \pmod{b/d}$; but due to the restriction on the range of possible values of $m,m'$, that further implies that $m=m'$ and therefore $n=n'$ also.
(Note that this argument assumes that $0 \notin \mathbb{N}$ in your convention, so that $a \ne 0$ and $b \ne 0$. If either or both of $a,b=0$, then those special cases should be easier to work with.)
Best Answer
For every $f(x,y)$ we can write $f(x,y)= (x^3-y^2)g(x,y)+ x^2a(y)+xb(y)+c(y)$. If $f(x,y)\in\ker\phi$, we have $z^4a(z^3)+z^2b(z^3)+c(z^3)=0$. Note that powers of terms in $c(z^3)$ are equal $0$ modulo $3$, powers of terms in $z^2b(z^3)$ are equal $2$ modulo $3$, and powers of terms in $z^4a(z^3)$ are equal $1$ modulo $3$. This implies that there are no cancelations in $z^4a(z^3)+z^2b(z^3)+c(z^3)$, so coefficients of all polynomials $a$, $b$ and $c$ are $0$, i.e. $a=b=c=0$. Thus $f(x,y)=(x^3-y^2)g(x,y)$ and $f(x,y)\in \langle x^3-y^2\rangle$.