How to prove $f(x) = 4x^{3} + 4x - 6$ has exactly one real root?

Your intuition for the first one is correct!

$f(0)<0$ and $f(1)>0$, so by IVT $f$ has a root say $x_0$

Suppose $f$ has another root $x_1 \neq x_0$ with $x_0<x_1$ .Then $f(x_0)=f(x_1)=0$ and by Rolles theorem $\exists$ $c \in (x_0,x_1)$ such that $f'(c)=0$, contradicting to the fact $f'(x)>0$

$f(x)=4x^3+4x-6$

$f(0)=-6$ and $f(-1)=-14$

By IVT, there exists at least one real root, $x\in(-1,0)$ such that $f(x)=0$

Now try to prove by using contradiction.

If not, there exists al least $2$ real roots $x_1,x_2$, such that $f(x_1)=f(x_2)=0$

Since $f(x)$ is differentiable, by using Rolle's Theorem, there exists a number $k\in(x_1,x_2)$ such that $f^{\prime}(k)=0$. But $f^{\prime}(x)=12x^2+4>0\ \forall x\ne0$

Your $f'(x)$ is strictly positive which means your function is strictly increasing. A strictly increasing function does not have more than one real root. Because otherwise it is not going to be one-to-one. Simply put, with more than one real root you have to have a turning point somewhere between those roots which makes your function both increasing and decreasing between the roots.