Following my nose gave the following argument. Writing $I$ be the left ideal generated by $x\partial^2$ and $x^3$ and using $\cdot$ to stress multiplication we get
$$ x^2 \cdot x\partial^2 - \partial^2\cdot x^3 = [ x^3, \partial^2] = -6x^2\partial - 6x\in I$$
So $$\frac{1}{6}x\partial\cdot (-6x^2\partial - 6x) + x^2\cdot x\partial^2 = x[x^2,\partial]\partial - x^2\partial + x[x,\partial] = -3x^2\partial - x\in I$$
Taking a suitable $\mathbb{k}$-linear combinations of these gives $x\in I$ and $x^2\partial\in I$.
Then $$\partial^2\cdot x - 1\cdot x\partial^2 = [x,\partial^2]=-2x\partial \in I.$$ Since also $2\partial x\in I$ we conclude $$\partial x- x\partial=[\partial,x]=1\in I$$ as required.
The general strategy at each step is to take two elements in the left ideal with the same principal symbol with respect to the filtration with $F_0=\mathbb{k}$, $F_1=\mathbb{k}\cdot\{1,x,\partial\}$ and $F_n=F_1^n$ and then compute their difference which will live in a lower filtered part. I haven't read the book but I'd imagine that this idea is at the heart of computations throughout.