Let's go ;D
\(f(y)=\frac{y^2}{y^3+8}\)
we have to use the quotient rule.
\(f(y)=\frac{g(y)}{h(y)}\)
\(f'(y)=\frac{h(y)*g'(y)-g(y)*h'(y)}{[h(y)]^2}\)
Then
\(g(y)=y^2\)
\(g'(y)=2y\)
\(h(y)=y^3+8\)
\(h(y)=3y^2\)
Now we can replace
\(f'(y)=\frac{h(y)*g'(y)-g(y)*h'(y)}{[h(y)]^2}\)
\(f'(y)=\frac{(y^3+8)*2y-(y^2)*3y^2}{(y^3+8)^2}\)
\(f'(y)=\frac{2y^4+16y-3y^4}{(y^3+8)^2}\)
\(\boxed{\boxed{f'(y)=\frac{16y-y^4}{(y^3+8)^2}}}\)
