Calc2.25

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x+xy+x^{2}+xy^{2}=0\,

You could solve this for y\, as it is a quadratic in y\, but it is probably much nicer to differentiate implicitly.

1+y+xy+2x+y^{2}+2xyy'=0'\,

Solving this for y' gives

y'={\frac  {-(1+y+2x+y^{2})}{x+2xy}}\,