To show that the function $v = \frac{-y}{x^2 + y^2}$ is harmonic, we need to verify that it satisfies the Laplace equation:

$\Delta v = \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} = 0$

Step 1: Compute the partial derivatives of $v$

We begin by computing the first and second partial derivatives of $v$ with respect to $x$ and $y$.

1. First partial derivative of $v$ with respect to $x$:

$v = \frac{-y}{x^2 + y^2}$ Using the quotient rule, we get: $\frac{\partial v}{\partial x} = \frac{(x^2 + y^2)(0) - (-y)(2x)}{(x^2 + y^2)^2} = \frac{2xy}{(x^2 + y^2)^2}$

2. First partial derivative of $v$ with respect to $y$:

Similarly, using the quotient rule: $\frac{\partial v}{\partial y} = \frac{(x^2 + y^2)(-1) - (-y)(2y)}{(x^2 + y^2)^2} = \frac{-(x^2 - y^2)}{(x^2 + y^2)^2}$

Step 2: Compute the second derivatives of $v$

Now, compute the second partial derivatives.

1. Second partial derivative of $v$ with respect to $x$: Differentiate $\frac{\partial v}{\partial x} = \frac{2xy}{(x^2 + y^2)^2}$ with respect to $x$ again, using the quotient rule: $\frac{\partial^2 v}{\partial x^2} = \frac{2y[(x^2 + y^2)^2] - 2xy(2(x^2 + y^2)2x)}{(x^2 + y^2)^4}$

Simplifying this expression is tedious, but it results in:

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

2. Second partial derivative of $v$ with respect to $y$: Differentiate $\frac{\partial v}{\partial y} = \frac{-(x^2 - y^2)}{(x^2 + y^2)^2}$ with respect to $y$: [ \frac{\partial^2 v}{\partial y^2} = \frac{-2y(x^2 - y^2) + 2y(x^2 + y^2)^2}{(x