Using chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
I'm given that $y=u^3+1$ and $u=5-x^2$
So far, I have came to $\frac{d}{du}(u^3+1) \cdot \frac{d}{dx}(5-x^2) = (3u^2)(-2x)$
Then: $(3u^2)(-2x) = -2x(3(5-x^2)^2)$
I'm not sure how to solve further, I can get $-6x$, but how to I deal with the squared part?
This is normally fine, however if you really do want to expand it then you can do it this way:
$$-2x(3(5-x^2)^2)=-6x(5-x^2)^2$$
Then, using the fact that $(a-b)^2=a^2-2ab+b^2$, we obtain:
$$-6x(5^2-2\cdot 5\cdot x^2+x^4)=-6x(25-10x^2+x^4)=-150x+60x^3-6x^5$$