1. **State the problem:** Solve the exact differential equation $$\left(\sin(x)\cos(y)+e^{2x}\right)dx + \left(\cos(x)\sin(y)+\tan(y)\right)dy = 0.$$\n\n2. **Check if the equation is exact:** Let \(M = \sin(x)\cos(y) + e^{2x}\) and \(N = \cos(x)\sin(y) + \tan(y)\).\nCalculate partial derivatives:\n$$\frac{\partial M}{\partial y} = -\sin(x)\sin(y), \quad \frac{\partial N}{\partial x} = -\sin(x)\sin(y).$$\nSince $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x},$$ the equation is exact.\n\n3. **Find the potential function \(\psi(x,y)\):**\nIntegrate \(M\) with respect to \(x\):\n$$\psi(x,y) = \int \left(\sin(x)\cos(y) + e^{2x}\right) dx = -\cos(x)\cos(y) + \frac{e^{2x}}{2} + h(y),$$ where \(h(y)\) is an unknown function of \(y\).\n\n4. **Differentiate \(\psi(x,y)\) with respect to \(y\) and equate to \(N\):**\n$$\frac{\partial \psi}{\partial y} = \cos(x)\sin(y) + h'(y) = \cos(x)\sin(y) + \tan(y).$$\nSo,\n$$h'(y) = \tan(y).$$\n\n5. **Integrate \(h'(y)\) to find \(h(y)\):**\n$$h(y) = \int \tan(y) dy = -\ln|\cos(y)| + C.$$\n\n6. **Write the implicit solution:**\n$$\psi(x,y) = -\cos(x)\cos(y) + \frac{e^{2x}}{2} - \ln|\cos(y)| = C,$$ where \(C\) is a constant.\n\n**Final answer:**\n$$-\cos(x)\cos(y) + \frac{e^{2x}}{2} - \ln|\cos(y)| = C.$$