Let us examine each option by plugging in $x$ and $y$ to determine if $f(x + y) = f(x)f(y)$ is satisfied and then assess their monotonicity.

Option A:
For $f(x) = x^3$, we find $f(x + y) = (x + y)^3$. Expanding $(x + y)^3$ does not result in a simple product of $x^3$ and $y^3$, thus $f(x + y) \neq f(x)f(y)$ and option A is incorrect.

Option B:
For $f(x) = 3^x$, let's check the functional equation:
$$f(x + y) = 3^{x+y} = 3^x \cdot 3^y = f(x)f(y),$$
which satisfies $f(x + y) = f(x)f(y)$. Also, since the base is greater than 1, $f(x) = 3^x$ is indeed a monotonically increasing function. Hence, option B is correct.

Option C:
For $f(x) = x^{\frac{1}{2}}$, the function is defined for $x \ge 0$. We have $f(x + y) = (x + y)^{\frac{1}{2}}$ which, even if it could be expressed as a product of the square roots of $x$ and $y$, will not satisfy the condition $f(x + y) = f(x)f(y)$ for all nonnegative $x$ and $y$. Therefore, option C is incorrect.

Option D:
For $f(x) = \left(\frac{1}{2}\right)^x$, we find $f(x + y) = \left(\frac{1}{2}\right)^{x+y}$. This results in a product of the form $\left(\frac{1}{2}\right)^x \left(\frac{1}{2}\right)^y$, which is a valid representation for $f(x)f(y)$, so the functional equation holds. However, this function is actually monotonically decreasing because the base is less than 1. Thus, option D is incorrect.

As options A, C, and D are incorrect and B is correct, we conclude that:
$$\boxed{f(x) = 3^x}$$
is the monotonically increasing function that satisfies $f(x + y) = f(x)f(y)$.