X^3-2y^3+xy(2x-y)+y(x-y)+1=0
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Let us solve for the derivative w.r.t xxx :
Let f(x,y)=x3−2y3+xy(2x−y)+y(x−y)+1f(x,y)=x^3-2y^3+xy(2x-y)+y(x-y)+1\\f(x,y)=x3−2y3+xy(2x−y)+y(x−y)+1
=x3−2y3+2x2y−xy2+xy−y2+1=x^3-2y^3+2x^2y-xy^2+xy-y^2+1=x3−2y3+2x2y−xy2+xy−y2+1
fx=∂dx[x3−2y3+2x2y−xy2+xy−y2+1]=3x2+4xy−y2+yf_x=\frac{\partial}{dx}[x^3-2y^3+2x^2y-xy^2+xy-y^2+1] \\ =3x^2+4xy-y^2+yfx=dx∂[x3−2y3+2x2y−xy2+xy−y2+1]=3x2+4xy−y2+y
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