How can I calculate the derivative of $g(z)=\int_{z^2}^{z^3}\frac{1}{\sqrt{1+t^2}} dt$?

$$f:t\mapsto \frac {1}{\sqrt {1+t^2}} $$ is continuous at R, thus

$$F:x\mapsto \int_0^xf (t)dt $$ is differentiable at R and $$F'(x)=f (x) $$

your function is $$g (z)=F (z^3)-F (z^2) $$

it is differentiable at R and

$$g'(z)=3z^2f (z^3)-2zf (z^2) $$

$$=\frac {3z^2}{\sqrt {1+z^6}}-\frac {2z}{\sqrt {1+z^4}} $$

Apply this: Differentiation under the integral sign.

$$\frac{\text{d}}{\text{d}x}\left( \int_{a(x)}^{b(x)}f(x,t)\text{d}t \right ) = f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)}^{b(x)}f_x(x,t)\text{d}t.$$

differetiating with respect to $x$ gives $$-2\,{\frac {x}{\sqrt {{x}^{4}+1}}}+3\,{\frac {{x}^{2}}{\sqrt {{x}^{6}+ 1}}} $$