Answer:[tex]\frac{dy}{dx}=\frac{4y-2x}{y-4x}[/tex]Step-by-step explanation:[tex]\frac{d}{dx}(2x^2)=4x[/tex][tex]\frac{d}{dx}(y^2)=2y\frac{dy}{dx}[/tex][tex]\frac{d}{dx}(8xy)[/tex][tex]=8\frac{d}{dx}(xy)[/tex][tex]=8(\frac{d}{dx}(x)y+x\frac{d}{dx}(y))[/tex][tex]=8[1y+x\frac{dy}{dx}][/tex][tex]=8y+8x\frac{dy}{dx}[/tex]Let's put it altogether now:[tex]2x^2+y^2=8xy[/tex]Differentiating both sides gives:[tex]4x+2y\frac{dy}{dx}=8y+8x\frac{dy}{dx}[/tex]We are solving for dy/dx so we need to gather those terms on one side and the terms without on the opposing side:I'm going to first subtract 4x on both sides:[tex]2y\frac{dy}{dx}=8y-4x+8x\frac{dy}{dx}[/tex]I'm not going to subtract 8xdy/dx on both sides:[tex]2y\frac{dy}{dx}-8x\frac{dy}{dx}=8y-4x[/tex]It is time to factor the dy/dx out of the two terms on the left:[tex]\frac{dy}{dx}(2y-8x)=8y-4x[/tex]Divide both sides by (2y-8x):[tex]\frac{dy}{dx}=\frac{8y-4x}{2y-8x}[/tex]Reduce right hand side fraction:[tex]\frac{dy}{dx}=\frac{4y-2x}{y-4x}[/tex]