Why $k[[x^2,x^3]]$ is dimension 1 obvious?

Let $f\in k[[x^2,x^3]]$ be nonzero and let $n$ be the smallest power of $x$ appearing in $f$. Note then that $f$ divides $x^{n+2}$ in $k[[x^2,x^3]]$: the usual procedure to divide by solving for the coefficients one by one will not require a linear term. So if $P\subset k[[x^2,x^3]]$ is a nonzero prime ideal, applying this to some nonzero element of $P$ gives that $x^{n+2}\in P$. Writing $x^{n+2}$ as a product of $x^2$'s and $x^3$'s this implies either $x^2$ or $x^3$ is in $P$, and then the other must be as well since $x^6$ is.

Set $f(Y):=Y^2-x^2$. Hence, $f(Y)$ is a monic polynomial in the polynomial ring $k[[x^2,x^3]][Y]$(the polynomial ring over $k[[x^2,x^3]]$). Now since $f(x)=0$, $k[[x]]$ is integral over $k[[x^2,x^3]]$. Thus, $\dim(k[[x^2,x^3]])=\dim(k[[x]])$. On the other hand, since $k[[x]]$ is a PID (it is a discrete valuation ring), it is 1 dimensional. Hence, $\dim(k[[x^2,x^3]])=\dim(k[[x]])=1$.