For the binomial expression $$\left[x^3-\frac{2}{x^2}\right]^6$$ I calculated the general term to be $$_6C_r x^{18-5r}(-2)^r$$
So for the term independent of $x$ , the exponent $18-5r=0$ but that would make $r=3.6$. I'm not too sure where I went wrong.
As you did, we have
$$\left(x^3-\frac{2}{x^2}\right)^6=\sum_{r=0}^{6}\binom 6r (x^3)^{6-r} \left(-\frac{2}{x^2}\right)^r=\sum_{r=0}^{6}\binom 6r(-2)^rx^{18-5r}$$
So, you did nothing wrong. Since there is no integer $r$ such that $18-5r=0$, there is no term independent of $x$.