What is the constant term in the expansion of ${\left(x^2-\dfrac{1}{x^4}\right)}^{13}$ ? and the middle term in it?

Question

I want to determine the constant term and middle term in the expansion of ${\left(x^2-\dfrac{1}{x^4}\right)}^{13}$, I know that the constant term in the expansion of $(x+a)^n$ is always $a^n $, We can see that the general term becomes constant when the exponent of variable $x$ is $0$ ,Therefore the condition for the constant term is $n-2k=0\implies k=\frac n2$, Really I can't use that fact to determine the constant term in the expansion of ${\left(x^2-\dfrac{1}{x^4}\right)}^{13}$ since both terms depend of $x$ however if we assume that $y=\dfrac 1x $ this means that y can be expressed as a ratio using the variable $x $ , In the expansion of ${\left(x^2-\dfrac{1}{x^4}\right)}^{13}$ the constant term according to its definition in the expansion of $(x+a)^n$ would be $-\dfrac{1287}{x^4}$ or $x^{26}$ the latter mixed me then What is the constant term in the expansion ${(x^2-\dfrac{1}{x^4})}^{13}$ ? and the middle term in it ?

Answer 1

$$\binom{13}{r}(x^2)^{13-r}\times (-x^{-4})^{r}=\binom{13}{r}x^0$$

$$26-2r-4r=0$$

$$3r=13$$

$r \notin Z$, there is no constant term.