Step 1: Use the link between analytic functions and harmonic functions.
If $g(z)$ is an analytic (holomorphic) function of $z = x+iy$, then $g(z)$ treated as a two dimensional function is automatically harmonic, because $\partial^2/\partial x^2 + \partial^2/\partial y^2$ acting on any power or exponential of $z$ picks up a factor of $1^2 + i^2 = 0$. This gives a quick way to spot harmonic functions without repeating the derivative calculation for every option.
Step 2: Test option (C) this way.
$f = e^{x+iy} = e^{z}$ with $z=x+iy$. This is exactly the function $e^{z}$, which is analytic everywhere (entire), so by Step 1 its Laplacian is zero automatically. No further calculation is needed for (C).
Step 3: Test option (D) by splitting it into known harmonic pieces.
Write $f = yx^{2} - \dfrac{y^{3}}{3} - xy$ as $f = \left(x^2y - \dfrac{y^3}{3}\right) - xy$. Now $z^{3} = (x+iy)^3 = x^3 - 3xy^2 + i(3x^2y - y^3)$, so $\text{Im}(z^3) = 3x^2y - y^3$, which means $x^2y - \dfrac{y^3}{3} = \dfrac{1}{3}\text{Im}(z^3)$, a harmonic function by Step 1. Also $z^2 = (x+iy)^2 = x^2 - y^2 + 2ixy$, so $\text{Im}(z^2) = 2xy$, meaning $xy = \dfrac{1}{2}\text{Im}(z^2)$, also harmonic by Step 1. A sum of two harmonic functions is harmonic since the Laplacian is linear, so $f$ is harmonic.
Step 4: Show (A) and (B) cannot be written this way.
For (A), $xe^{y} - ye^{x}$ mixes a polynomial in one variable with an exponential in the other in a way that does not match the real or imaginary part of any elementary analytic function of $z=x+iy$; a direct check gives $f_{xx}+f_{yy} = xe^y - ye^x \neq 0$. For (B), $x\cos(y) - y\cos(x)$ has the same mismatch, and a direct check gives $f_{xx}+f_{yy} = y\cos(x) - x\cos(y) \neq 0$. Neither is harmonic.
Final Answer:
Only $e^{x+iy}$ and $yx^2 - \dfrac{y^3}{3} - xy$ are harmonic.
\[ \boxed{e^{x+iy} \text{ and } yx^{2} - \dfrac{y^{3}}{3} - xy} \]