$\text{ If }xf\left(x\right)=3f^2\left(x\right)+2 \text{ Then }\int \frac{2x^2-12xf\left(x\right)+f\left(x\right)}{\left(6f\right(x)-x){\left(x^2-f\left(x\right)\right)}^2}dx=$

$xf\left(x\right)=3f^2\left(x\right)+2$

$\begin{array}{l}\text{ Differentiating both sides wrtx }\\ x{f}^{\mathrm{\prime }}\left(x\right)+f\left(x\right)=6f\left(x\right)f^{\mathrm{\prime }}\left(x\right)\\ x{f}^{\mathrm{\prime }}\left(x\right)-6f\left(x\right)f\left(x\right)=-f\left(x\right)\\ \left(x-6f\left(x\right)f^{\mathrm{\prime }}\left(x\right)=-f\left(x\right)\right.\\ f^{\mathrm{\prime }}\left(x\right)=\frac{-f\left(x\right)}{x-6f\left(x\right)}\\ f^{\mathrm{\prime }}\left(x\right)=\frac{f\left(x\right)}{6f\left(x\right)-x}----\left(1\right)\end{array}$

$\begin{array}{r}\text{ Now }=\int \frac{2x^2-12xf\left(x\right)+f\left(x\right)}{\left(6f\right(x)-x){\left(x^2-f\left(x\right)\right)}^2}dx\\ =\int \frac{2x(x-6f(x\left)\right)+f\left(x\right)}{\left(6f\right(x)-x){\left(x^2-f\left(x\right)\right)}^2}dx\end{array}$

$\begin{array}{l}=\int \frac{-2x}{{\left(x^2-f\left(x\right)\right)}^2}dx+\int \frac{f\left(x\right)}{\frac{f\left(x\right)}{f^{\left(x\right)}}{\left(x^2-f\left(x\right)\right)}^2}dx\\ =\int \frac{-2x}{{\left(x^2-f\left(x\right)\right)}^2}dx+\int \frac{f^{\mathrm{\prime }}\left(x\right)}{{\left(x^2-f\left(x\right)\right)}^2}dx\\ =\int \frac{-2x-f^{\mathrm{\prime }}\left(x\right)}{{\left[x^2-f\left(x\right)\right]}^2}dx\\ =\int \frac{1}{x^2-f\left(x\right)}+C\\ \text{ Put }{x}^2-f\left(x\right)=t\\ \left(2x-f^{\mathrm{\prime }}\left(x\right)dx=dt\right)\\ =-\int \frac{1}{t^2}dt=\frac{1}{t}+C\\ =\frac{1}{x^2-f\left(x\right)}+C\end{array}$