Treat #y# as a function of #x# and use the Chain Rule whenever #y# appears. See here for more details on implicit differentiation.
Here, differentiating both sides with respect to #x# and taking the Product Rule and Chain Rule into account yields
#\frac{dy}{dx}\cos(x^2)-2xy\sin(x^2)-2y\frac{dy}{dx}=y+x\frac{dy}{dx}-2x#
Put all terms that contain #\frac{dy}{dx}# on one side, all the other terms on the other sides:
#\frac{dy}{dx}\cos(x^2)-2y\frac{dy}{dx}-x\frac{dy}{dx}=2xy\sin(x^2)+y-2x#
Factoring #\frac{dy}{dx}# and dividing, we obtain
#\frac{dy}{dx}=\frac{2xy\sin(x^2)+y-2x}{\cos(x^2)-2y-x}#
