Answer
$$\frac{{dy}}{{dx}} = \frac{{7{y^2} - 3{x^2} - 4x{y^2} - 4{x^3}}}{{2y\left( {2{x^2} + 2{y^2} - 7x} \right)}}$$
Work Step by Step
$$\eqalign{
& \left( {{x^2} + {y^2}} \right)\left( {{x^2} + {y^2} + x} \right) = 8x{y^2} \cr
& {\text{Distributive property}} \cr
& {x^4} + {x^2}{y^2} + {x^3} + {x^2}{y^2} + {y^4} + x{y^2} = 8x{y^2} \cr
& {x^4} + 2{x^2}{y^2} + {x^3} + {y^4} + x{y^2} = 8x{y^2} \cr
& {x^4} + 2{x^2}{y^2} + {x^3} + {y^4} - 7x{y^2} = 0 \cr
& {\text{Differentiate both sides with respect to }}x \cr
& 4{x^3} + 4{x^2}yy' + 4x{y^2} + 3{x^2} + 4{y^3}y' - 14xyy' - 7{y^2} = 0 \cr
& {\text{Collect like terms}} \cr
& 4{x^2}yy' + 4{y^3}y' - 14xyy' = 7{y^2} - 3{x^2} - 4x{y^2} - 4{x^3} \cr
& \left( {4{x^2}yy' + 4{y^3}y' - 14xyy'} \right) = 7{y^2} - 3{x^2} - 4x{y^2} - 4{x^3} \cr
& \left( {4{x^2}y + 4{y^3} - 14xy} \right)y' = 7{y^2} - 3{x^2} - 4x{y^2} - 4{x^3} \cr
& {\text{Solve for }}y' \cr
& \frac{{dy}}{{dx}} = \frac{{7{y^2} - 3{x^2} - 4x{y^2} - 4{x^3}}}{{4{x^2}y + 4{y^3} - 14xy}} \cr
& \frac{{dy}}{{dx}} = \frac{{7{y^2} - 3{x^2} - 4x{y^2} - 4{x^3}}}{{2y\left( {2{x^2} + 2{y^2} - 7x} \right)}} \cr} $$
