The function $\frac{x^3}{3}+2x^2$ does not grow exponentially faster than $x^2+4x$. The function $\frac{x^3}{3}+2x^2$ is a polynomial, and it grows roughly like $(x^2+4x)^{3/2}$. That is not dramatically faster than $x^2+4x$. The following is an example of faster relative growth that is more familiar.
Look at the area of the region under the curve $y=x$, above the $x$-axis, from $x=0$ to $x=w$. If you draw the "curve" $y=x$, you will see that the region is a triangle with base $w$ and height $w$, so the region has area $w^2/2$. (The same result can be obtained by integration, but that's overkill.)
So the function grows like $w$, and the area grows like $w^2$, more precisely like a constant times $w^2$.
The most "extreme" example is a constant function like $f(x)=1$. The area from $0$ to $w$ under the curve is $w$. This grows much faster than $f$, which does not grow at all, but the rate of growth of the area would not be considered fast. And the fact that the area of a rectangle of constant height grows linearly with the base is unsurprising.
In general, imagine a curve $y=f(x)$, where $f(x)$ is positive, and, for simplicity, increasing. Let $W$ be the region below $y=f(x)$, above the $x$-axis, from $x=0$ to $x=w$. Then the region $W$ is contained in the rectangle with base the interval from $0$ to $w$, and height $f(w)$. So the area of $W$ is $\le wf(w)$. Thus, if $f(w)$ grows "fast," then the rate of growth of the area is not much faster than the rate of growth of $f(w)$.
Consider the differential equation
$$
\frac{f'}{f} = g
$$
where $g$ is the fast-growing function you want. For instance, for $g(x) = e^x$ (and say the initial condition $f(0) =1$) you get
$$f(x) = e^{e^x-1}
$$
The ratio $f'/f$ grows arbitrarily large.
Best Answer
Compute the roots of $h(x)=f(x)-g(x)$.
Then you know what intervalls are interesting.
Those intervalls are a finite number for most excercises you encounter, or have a very easy to recognize pattern (like sin(x)-cos(x) ).
Then check with values inside those intervalls, wether f(x)>g(x) or vice versa.
Can you explain why this works?