Consider the function
z(x,y)=y3+2xy2−3x2y.
Determine the tangent line through (x,y)=(2,1).
z′x(x,y)z′y(x,y)=2y2−6xy3y2+4xy−3x2
slope=−2⋅12−6⋅2⋅13⋅12+4⋅2⋅1−3⋅22=−−10−1=−10.
The general form of a tangent line is t(x)=ax+b. Now it holds that a=−10.
In order to determine b we use that the tangent line goes through (x,y)=(2,1). Hence, t(2)=−10⋅2+b=1, which results in b=21.
Hence, the tangent line is given by t(x)=−10x+21.