$XY\in(XY)$, but neither $X\in(XY)$ nor $Y\in (XY)$, hence $(XY)$ is not prime (and even less maximal).
$(X+Y)$ is the kernel of the homomorphism $\phi\colon\mathbb C[X,Y]\to \mathbb C[X]$ induced by $X\mapsto X$, $Y\mapsto -X$ and $\mathbb C[X]$ has no zero-divisors (but is not a field). Hence $(X+Y)$ is prime (but not maximal).
To see that indeed $\ker\phi=(X+Y)$ (with $\supseteq$ being trivial), consider a nonzero element $f(X,Y)\in\ker\phi$ of minimal degree $n$ in $Y$. Collect all monomials of same degree in $Y$ and thus write $f(X,Y) = g_0(X)+Yg_1(X)+ \ldots + Y^{n-1}g_{n-1}(X)+ Y^ng_n(X).$ If $n=0$, then $\phi(f)=g_0=0$ implies $f=0$. Therefore $n\ge1$. If we add a multiple of $X+Y$ to $f$, we obtain another element of $\ker\phi$, for example $f(X,Y)-(X+Y)Y^{n-1}g_n(X) = g_0(X)+Yg_1(X)+ \ldots + Y^{n-1}(g_{n-1}(X)-Xg_n(X)).$ As this has at most degree $n-1$ in $Y$ it must be the zero polynomial by minimality of $n$, that is $g_0=\cdots=g_{n-2}=0$ and $g_{n-1}=Xg_n$. But then $f(X,Y)=Y^{n-1}(X+Y)g_n(X)$ is a multiple of $X+Y$.