1. **State the problem:** We are given a harmonic function $f(x,y)$ and need to show that the function $$g(x,y) = f\left(\frac{x}{x^2+y^2}, -\frac{y}{x^2+y^2}\right)$$ is also harmonic. 2. **Recall the definition of harmonic functions:** A function $u(x,y)$ is harmonic if it satisfies Laplace's equation: $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0.$$ 3. **Given:** $f$ is harmonic, so $$f_{xx} + f_{yy} = 0,$$ where subscripts denote partial derivatives with respect to those variables. 4. **Define:** $$X = \frac{x}{x^2+y^2}, \quad Y = -\frac{y}{x^2+y^2}.$$ We want to show that $$g(x,y) = f(X,Y)$$ is harmonic, i.e., $$g_{xx} + g_{yy} = 0.$$ 5. **Use the chain rule to find $g_x$ and $g_y$: ** $$g_x = f_X X_x + f_Y Y_x,$$ $$g_y = f_X X_y + f_Y Y_y,$$ where $f_X = \frac{\partial f}{\partial X}$ and $f_Y = \frac{\partial f}{\partial Y}$. 6. **Compute the first derivatives of $X$ and $Y$: ** Let $r^2 = x^2 + y^2$. $$X = \frac{x}{r^2}, \quad Y = -\frac{y}{r^2}.$$ Calculate: $$X_x = \frac{r^2 - 2x^2}{r^4} = \frac{y^2 - x^2}{r^4},$$ $$X_y = \frac{-2xy}{r^4},$$ $$Y_x = \frac{2xy}{r^4},$$ $$Y_y = \frac{-r^2 + 2y^2}{r^4} = \frac{y^2 - x^2}{r^4}.$$ 7. **Calculate second derivatives $g_{xx}$ and $g_{yy}$ using the product and chain rules:** $$g_{xx} = f_{XX} (X_x)^2 + 2 f_{XY} X_x Y_x + f_{YY} (Y_x)^2 + f_X X_{xx} + f_Y Y_{xx},$$ $$g_{yy} = f_{XX} (X_y)^2 + 2 f_{XY} X_y Y_y + f_{YY} (Y_y)^2 + f_X X_{yy} + f_Y Y_{yy}.$$ 8. **Sum $g_{xx} + g_{yy}$:** Group terms: $$g_{xx} + g_{yy} = f_{XX}[(X_x)^2 + (X_y)^2] + 2 f_{XY}[X_x Y_x + X_y Y_y] + f_{YY}[(Y_x)^2 + (Y_y)^2] + f_X (X_{xx} + X_{yy}) + f_Y (Y_{xx} + Y_{yy}).$$ 9. **Use harmonicity of $f$: ** Since $f$ is harmonic, $$f_{XX} + f_{YY} = 0.$$ 10. **Show that the coefficients satisfy:** $$(X_x)^2 + (X_y)^2 = (Y_x)^2 + (Y_y)^2,$$ $$X_x Y_x + X_y Y_y = 0,$$ $$X_{xx} + X_{yy} = 0,$$ $$Y_{xx} + Y_{yy} = 0.$$ 11. **Verify these identities:** - Compute $(X_x)^2 + (X_y)^2$ and $(Y_x)^2 + (Y_y)^2$: $$X_x = \frac{y^2 - x^2}{r^4}, \quad X_y = \frac{-2xy}{r^4},$$ so $$(X_x)^2 + (X_y)^2 = \frac{(y^2 - x^2)^2 + 4x^2 y^2}{r^8} = \frac{(x^2 + y^2)^2}{r^8} = \frac{r^4}{r^8} = \frac{1}{r^4}.$$ Similarly for $Y$: $$Y_x = \frac{2xy}{r^4}, \quad Y_y = \frac{y^2 - x^2}{r^4},$$ $$(Y_x)^2 + (Y_y)^2 = \frac{4x^2 y^2 + (y^2 - x^2)^2}{r^8} = \frac{r^4}{r^8} = \frac{1}{r^4}.$$ - Compute $X_x Y_x + X_y Y_y$: $$X_x Y_x + X_y Y_y = \frac{y^2 - x^2}{r^4} \cdot \frac{2xy}{r^4} + \frac{-2xy}{r^4} \cdot \frac{y^2 - x^2}{r^4} = 0.$$ - Compute $X_{xx} + X_{yy}$ and $Y_{xx} + Y_{yy}$: By direct differentiation, both equal zero (harmonicity of $X$ and $Y$ as functions). 12. **Substitute back:** $$g_{xx} + g_{yy} = f_{XX} \frac{1}{r^4} + 0 + f_{YY} \frac{1}{r^4} + f_X \cdot 0 + f_Y \cdot 0 = \frac{1}{r^4} (f_{XX} + f_{YY}) = 0,$$ since $f$ is harmonic. 13. **Conclusion:** The function $$g(x,y) = f\left(\frac{x}{x^2+y^2}, -\frac{y}{x^2+y^2}\right)$$ is harmonic. **Final answer:** $$\boxed{g \text{ is harmonic if } f \text{ is harmonic.}}$$