$ \frac{\partial v}{\partial x} = 3x^2 - 3y^2 $

$ \frac{\partial^2 v}{\partial x^2} = 6x $

$ \frac{\partial v}{\partial y} = - 6xy + 2 $

$ \frac{\partial^2 v}{\partial y^2} = - 6x $

Now, $ \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 6x - 6x = 0 $

Thus the given function $ v(x,y) $ is harmonic.

Let the harmonic conjugate be $ u(x,y) $. Then,

$ \frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y} \implies \frac{\partial u}{\partial y} = - 3x^2 + 3y^2 \implies u = -3x^2y + y^3 + c_1(x)$

Also, $ \frac{\partial v}{\partial y} = \frac{\partial u}{\partial x} \implies \frac{\partial u}{\partial x} = - 6xy + 2 \implies u = - 3x^2y + 2x + c_2(y)$

Hence, $u(x,y) = - 3x^2y + y^3 + 2x $

Thus, the analytic function with real and imaginary parts as $u$ and $v$ respectively is,

$f(u,v) = u + iv = (-3x^2y + y^3 + 2x) + i (x^3-3xy^2+2y) $