Derivative with respect to the integration limit returns the integrand, evaluated at the limit.
$$\frac{d}{dz}\int_3^{z}e^{t^3}dt=e^{z^3}$$ Because $z=x^2$ and you take the derivative with respect to $x$, you need the chain rule.
$$\frac{d}{dx}=\frac{dz}{dx}\frac{d}{dz}$$ so now you have $$\frac{d}{dx}\int_3^{x^2}e^{t^3}dt=\left(\frac{d(x^2)}{dx}\right)e^{x^6}=2xe^{x^6}$$
The derivative with respect to the lower limit is similar (remember that the result of integration is the indefinite integral, evaluated at both limits, with a minus between them). So in general, you have
$$\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)\,dt=f(b(x))b'(x)-f(a(x))a'(x)$$ If $f$ also depends on x, you have another term that involves derivative under the integral sign.