For what values of $a, b, c$ can $u(x,y)=ax^2+bxy+cy^2$ be the real part of an analytic function? Obviously $0$ works, but that's not the interesting case. I think we're to make use of the the CR equations, so naturally,
$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}=2ax+by$$
and
$$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}=bx+2cy$$
But with this, I'm not sure what to do. I know harmonic functions are not guarnanteed to be analytic; instead, satisfying the CR equations is a stronger property. I don't think integrating is useful here. I suppose we make the assumption the first derivative is $0$, and then just try and solve for one constant using both of these equations?
$$2ax=-by, bx=-2cy\implies x=-\frac{2cy}{b}\implies -2a(\frac{2cy}{b})=-by\implies$$
$$4ac=b^2\implies a=\frac{b^2}{4c}$$
but then we just get nested coefficients, so I don't think this is the right approach.
Update: Based off the comments, it would seem I need to solve Laplace's equation indeed. Then, we have
$$\frac{\partial^2u}{\partial x^2}=2a, \text{ and }\frac{\partial^2u}{\partial y^2}=2c\implies$$
$$\boxed{a=-c}$$
We then take the second partial of our $\partial v$ functions and totally omit having to know $v$ giving us
$$\frac{\partial^2v}{\partial y^2}=b, \text{ and }\frac{\partial^2v}{\partial x^2}=-b\implies$$ $$b=b\implies \boxed{b\in\mathbb{R}}$$
Let assume that $u$ is a real part of an holomorphic function, then $u$ must be harmonic, i.e, $u$ must satisfy $u_{xx}+u_{yy}=0$, $$\frac{\partial u}{\partial x}=2ax+by \Rightarrow u_{xx}=2a$$
Similarly, $u_{yy}=2c$. Thus $u$ is harmonic when $a=-c$ and $b$ any real number. This means that $u$ is a real part of an holomorphic function $f=u+iv$, where $v$ can be computed from the Cauchy-Riemann equation, when $a=-c$ and $b$ is real.