The first equation, \(\displaystyle \left(\frac{x+ y}{x}\right)dx+ (ln(x)+ y^4+ 3y^2+ 5)dy= 0\) is an "exact" equation, \(\displaystyle \frac{\partial \left(\frac{x+ y}{x}\right)}{\partial y}= \frac{1}{x}= \frac{\partial (ln(x)+ y^4+ 3y^2+ 5)}{\partial x}\).
That means that there exist a function F(x, y), such that \(\displaystyle \frac{\partial F}{\partial x} = \frac{x+ y}{x}\) and \(\displaystyle \frac{\partial F}{\partial y}= ln(x)+ y^4+ 3y^2+ 5\).
From the first, since the partial derivative with respect to x treats y as a constant, we can integrate treating y as a constant. \(\displaystyle \int \frac{x+ y}{x}dx= \int 1+ \frac{y}{x} dx= x+ y ln(x)+ C\) except that the constant, C, might actually be a function of y: \(\displaystyle F(x, y)= x+ y ln(x)+ g(y)\).
Now we have \(\displaystyle \frac{\partial F}{\partial y}= ln(x)+ \frac{dg}{dx}= ln(x)+ y^4+ 3y^2+ 5\).
Finish it.