\(\displaystyle \L\\x^{2}y^{2}-2xy=0\)
\(\displaystyle \L\\x^{2}2y\frac{dy}{dx}+2xy^{2}-(2x\frac{dy}{dx}+2y)=0\)
\(\displaystyle \L\\\frac{dy}{dx}=\frac{-2xy^{2}+2y}{2x^{2}y-2x}\)
\(\displaystyle \L\\\frac{-2y\sout{(xy-1)}}{2x\sout{(xy-1)}}=\frac{-y}{x}\)
Now, differentiate -y/x, quotient rule:
\(\displaystyle \L\\\frac{x\overbrace{(\frac{-dy}{dx})}^{\text{Remember,\\dy/dx=-y/x}}-(-y)}{x^{2}}=\frac{-x(\frac{-y}{x})+y}{x^{2}}=\frac{2y}{x^{2}}\)
