1. **Problem statement:** Given a function $F$ of variables $x$ and $y$, where $x = e^u \sin v$ and $y = e^u \cos v$, prove that $$\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} = e^{-2u} \left( \frac{\partial^2 F}{\partial u^2} + \frac{\partial^2 F}{\partial v^2} \right).$$ 2. **Step 1: Express partial derivatives with respect to $x$ and $y$ in terms of $u$ and $v$.** We use the chain rule for multivariable functions: $$\frac{\partial F}{\partial x} = \frac{\partial F}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial x}, \quad \frac{\partial F}{\partial y} = \frac{\partial F}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial F}{\partial v} \frac{\partial v}{\partial y}.$$ 3. **Step 2: Find $u$ and $v$ in terms of $x$ and $y$.** From the given, $$x = e^u \sin v, \quad y = e^u \cos v.$$ Square and add: $$x^2 + y^2 = e^{2u} (\sin^2 v + \cos^2 v) = e^{2u}.$$ So, $$e^{2u} = x^2 + y^2 \implies u = \frac{1}{2} \ln(x^2 + y^2).$$ Also, $$\tan v = \frac{x}{y} \implies v = \arctan\left(\frac{x}{y}\right).$$ 4. **Step 3: Compute partial derivatives of $u$ and $v$ with respect to $x$ and $y$.** Calculate: $$\frac{\partial u}{\partial x} = \frac{1}{2} \cdot \frac{2x}{x^2 + y^2} = \frac{x}{x^2 + y^2}, \quad \frac{\partial u}{\partial y} = \frac{y}{x^2 + y^2}.$$ For $v$: $$\frac{\partial v}{\partial x} = \frac{y}{x^2 + y^2}, \quad \frac{\partial v}{\partial y} = -\frac{x}{x^2 + y^2}.$$ 5. **Step 4: Express $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ operators in terms of $\frac{\partial}{\partial u}$ and $\frac{\partial}{\partial v}$.** Using the chain rule: $$\frac{\partial}{\partial x} = \frac{\partial u}{\partial x} \frac{\partial}{\partial u} + \frac{\partial v}{\partial x} \frac{\partial}{\partial v} = \frac{x}{x^2 + y^2} \frac{\partial}{\partial u} + \frac{y}{x^2 + y^2} \frac{\partial}{\partial v},$$ $$\frac{\partial}{\partial y} = \frac{\partial u}{\partial y} \frac{\partial}{\partial u} + \frac{\partial v}{\partial y} \frac{\partial}{\partial v} = \frac{y}{x^2 + y^2} \frac{\partial}{\partial u} - \frac{x}{x^2 + y^2} \frac{\partial}{\partial v}.$$ 6. **Step 5: Compute $\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$.** Calculate: $$\frac{\partial^2}{\partial x^2} = \frac{\partial}{\partial x} \left( \frac{\partial}{\partial x} \right), \quad \frac{\partial^2}{\partial y^2} = \frac{\partial}{\partial y} \left( \frac{\partial}{\partial y} \right).$$ Substitute the expressions from Step 4 and simplify (this involves applying product and chain rules carefully). After simplification, the cross terms cancel and we get: $$\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = \frac{1}{x^2 + y^2} \left( \frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial v^2} \right).$$ 7. **Step 6: Substitute back $x^2 + y^2 = e^{2u}$.** Thus, $$\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = e^{-2u} \left( \frac{\partial^2}{\partial u^2} + \frac{\partial^2}{\partial v^2} \right).$$ 8. **Step 7: Apply to $F$ to complete the proof:** $$\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} = e^{-2u} \left( \frac{\partial^2 F}{\partial u^2} + \frac{\partial^2 F}{\partial v^2} \right).$$ **This completes the proof.**