I think $\frac{d}{dx} \int f(x) dx = f(x)$ right? So $\frac{d}{dx} \int^b_a f(x) dx = [f(x)]^b_a = f(a)-f(b)$? But why when:
$$f(x) = \int^{x^3}_{x^2} \sqrt{7+2e^{3t-3}}$$
then
$$f'(x) = \color{red}{(x^3)'}\sqrt{7+2e^{3x-3}} - \color{red}{(x^2)'}\sqrt{7+2e^{3x-3}}$$
Where did the $(x^3)'$ and $(x^2)'$ come from?
On
For a definite integral with a variable upper limit of integration $\int_a^xf(t)\,dt$, you have ${d\over dx} \int_a^xf(t)\,dt=f(x)$.
For an integral of the form $$\tag{1}\int_a^{g(x)} f(t)\,dt,$$ you would find the derivative using the chain rule.
As stated above, the basic differentiation rule for integrals is:
$\ \ \ \ \ \ $for $F(x)=\int_a^x f (t)\,dt$, we have $F'(x)=f(x)$.
The chain rule tells us how to differentiate $(1)$. Here if we set $F(x)=\int_a^x f(t)\,dt$, then the derivative sought is $${d\over dx} \int_a^{g(x)} f(t)\,dt =[F(g(x))]' =F' (g(x)) g'(x) =f(g(x))\cdot g'(x).$$
So for example, given $$ {d\over dx} \int_0^{x^3} \sqrt{7+2e^{3t-3}}\, dt, $$ we have $F(x)=\int_0^x \sqrt {7+2e^{3t-3}}\,dt$, and we want to find the derivative of $F(x^3)$. Using the chain rule $$ {d\over dx} \int_0^{x^3}\underbrace{ \sqrt{7+2e^{3t-3}}}_{f(t)}\, dt = [F(x^3)]'=f(x^3)\cdot(x^3)'=\sqrt{7+3{e^{3x^3-3} }}\cdot 3x^2. $$ Note you have a mistake in the exponents in your solution.
If both the upper and lower limits of integration are variables, you'd do as you suggest. For example, you'd write $$\eqalign{ \int_{x^2}^{x^3}f(t)\,dt&= \int_{x^2}^0f(t)\,dt+ \int_0^{x^3}f(t)\,dt\cr &= -\int_{0}^{x^2}f(t)\,dt+ \int_0^{x^3}f(t)\,dt} $$ The derivative will then be, applying the chain rule to both integrals above $-f(x^2)\cdot2x+f(x^3)\cdot (3x^2)$.
$\int_a^bf(x)\,dx$ is a number, so ${d\over dx}\int_a^bf(x)\,dx=0$.
Now suppose $\int g(x)\,dx=F(x)$. Then $\int_{x^2}^{x^3}g(t)\,dt=F(x^3)-F(x^2)$, so ${d\over dx}\int_{x^2}^{x^3}g(t)\,dt=(x^3)'F'(x^3)-(x^2)'F'(x^2)=3x^2g(x^3)-2xg(x^2)$.