The function $F(x)$ is defined by the following integral
$$F(x)=\int_0^x\frac{\left(1-y^3\right)^a}{\sqrt{\left(\dfrac{1-y^3}{1-x^3}\right)^b-\left(\dfrac{y}{x}\right)^4}}\,dy$$
where $a$ and $b$ are constants. Depending on the values of $a$ and $b$, the function $F(x)$ can be divergent at $x=1$.
What is the asymptotic behavior of $F(x)$ in the $x\to 1$ limit?
When $F(x)$ is divergent at $x=1$, we only need to analyze the $y\approx x\approx 1$ part of the integral. The difficulty is that two limits are involved: $x\to 1$ and $y\to 1$.
You can get through a lot of the complication by scaling. For example, let $y=x u$; then
$$F(x) = x \int_0^1 du \frac{(1-x^3 u^3)^a}{\displaystyle \sqrt{\left (\frac{1-x^3 u^3}{1-x^3}\right )^b-u^4}} $$
It should be plain to see that
$$F(x) \sim x \int_0^1 \frac{du}{\sqrt{1-u^4}} = \frac{\Gamma^2\left ( \frac14 \right )}{4 \sqrt{2 \pi}} x \quad (x \to 0)$$
i.e., $F(x)$ vanishes as $x \to 0$.
To treat $x \to 1^-$, let $x=1-q$ and consider the limit as $q \to 0$. You will find that the integral is, to first order in $q$,
$$F(x) \sim (3 q)^{b/2} \int_0^1 du \, (1-u^3)^{a-b/2} = (3 (1-x))^{b/2} \frac{\Gamma \left ( \frac43\right ) \Gamma \left ( a-\frac{b}{2} + 1\right )}{\Gamma \left ( a-\frac{b}{2} + \frac43\right )} \quad (x \to 1^-)$$
which also clearly vanishes when $b \gt 0$. Higher orders may be computed by simple Taylor expansions.