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Find the general solution to the exact equation:

$$e^x\ln y + \left( \frac{e^x}{y} + 3y^2 \right) \frac{dy}{dx} = 1 \quad \text{for} \quad y > 0$$

$\Phi(x,y) = e^x \ln y + y^3 = 0$

$\Phi(x,y) = e^x \ln y + x + y^3 = 0$

$\Phi(x,y) = e^x \ln y - x + y^3 = 0$

$\Phi(x,y) = e^x \ln y - x + y^3 + C = 0$

There does not exist a general solution to the given exact equation.