Consider the ring $R[t]$, where $t$ is a variable. Since $R$ is an integral domain, so is $R[t]$.
Define the function $\Psi:R[x,y] \to R[t]$ sending $x$ to $t^2$ and $y$ to $t^3$, and extend it to for any $f(x,y)\in R[x,y]$ by linearity.
Then the map $\Psi$ is a ring homomorphism.

Our goal is to show that the kernel of $\Psi$ is the ideal $(x^3-y^2)$. If this is proved, then the first isomorphism theorem implies that $R[x,y]/(x^3-y^2)$ is a subring of the integral domain $R[t]$, and thus $(x^3-y^2)$ is a prime ideal.

Therefore it suffices to show that $\ker(\Psi)=(x^3-y^2)$.
Since we have
\[\Psi(x^3-y^2)=(t^2)^3-(t^3)^2=t^6-t^6=0,\]
one inclusion $(x^3-y^2) \subset \ker(\Psi)$ is clear.

To show that the other inclusion $\ker(\Psi) \subset (x^3-y^2)$, note that for any $f(x,y)\in R[x,y]$ we can write
\[f(x,y)=f_0(x)+f_1(x)y+(x^3-y^2)g(x,y), \tag{*}\]
where $f_0(x), f_1(x)\in R[x]$ and $g(x,y)\in R[x,y]$.
To obtain this expression, note that we have
\[x^my^n=-(x^3-y^2)x^my^{n-2}+x^{m+3}y^{n-2}\]
for integers $m$ and $n\geq 2$.

Use this relation recursively (to the second term), we can express
\[x^my^n=(x^3-y^2)p(x,y)+q_0(x)+q_1(x)y\]
for some $p(x,y)\in R[x,y]$ and $q_0, q_1\in R[x]$.
Hence we obtain the expression (*).

Suppose that $f(x,y)\in \ker(\Psi)$ and $f(x,y)$ can be written as in (*).
Then we have
\begin{align*}
0&=\Psi(f(x,y))=f(t^2,t^3)\\
&=f_0(t^2)+f_1(t^2)t^3+((t^2)^3-(t^3)^2)g(t^2, t^3)\\
&=f_0(t^2)+f_1(t^2)t^3.
\end{align*}

Note that the even terms of the polynomial $f_0(t^2)+f_1(t^2)t^3$ are $f_0(t^2)$, and the odd terms are $f_1(t^2)t^3$.
Thus, we have $f_0(t^2)=0$, and $f_1(t^2)=0$.
Equivalently, we have $f_0(x)=0$ and $f_1(x)=0$.

It follows that
\[f(x,y)=(x^3-y^2)g(x,y)\]
and it is in the ideal $(x^3-y^2)$.
Thus we prove that $\ker(\Psi)=(x^3-y^2)$ as required.

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