The general form for your differential equation is
$$M(x,y)dx+N(x,y)dy=0$$
Thus we identify:
$$M=3x+\dfrac{6}{y},\quad N=\dfrac{x^2}{y}+\dfrac{3y}{x}$$
We then compute the derivatives:
$$M_x=3,\quad M_y=-\dfrac{6}{y^2},\quad N_x=\dfrac{2x}{y}-\dfrac{3y}{x^2},\quad N_y=-\dfrac{x^2}{y^2}+\dfrac{3}{x}$$
The first thing to check, for exactness, is whether $M_y=N_x$. Clearly this isn't the case.
Next, we look at the following expression and check if it is a function of $x$ only:
$$\dfrac{M_y-N_x}{N}=\dfrac{-6x^2-2x^3y+3y^3}{x^2y^2}\cdot\dfrac{xy}{x^3+3y^2}\neq f(x)$$
Therefore the integrating factor is not a function only of $x$. We check similarly for $y$:
$$\dfrac{N_x-M_y}{M}=\dfrac{6x^2+2x^3y-3y^3}{x^2y^2}\cdot\dfrac{y}{3xy+6}\neq f(y)$$
Therefore the integrating factor is not a function only of $y$. We finally check for a function of both $x$ and $y$:
$$\dfrac{N_x-M_y}{xM-yN}=\dfrac{6x^2+2x^3y-3y^3}{x^2y^2}\cdot\dfrac{xy}{2x^3y+6x^2-3y^3}=\dfrac{1}{xy}=f(x,y)$$
Therefore, the integrating factor will be:
$$\mu(x,y)=\exp\left(\int \dfrac{dx}{x}\right)\exp\left(\int\dfrac{dy}{y}\right)=xy$$