Answer
The tangent line to the curve at $P(2,1)$ is $y=-11x+23$.
Work Step by Step
$$x^3-xy+y^3=7$$
The graph of the curve is enclosed below.
a) As the point $(2,1)$ lies in the curve, $P$ satisfies the equation.
b) Find $dy/dx$ using implicit differentiation: $$3x^2-(y+x\frac{dy}{dx})+3y^2\frac{dy}{dx}=0$$ $$3x^2-y-x\frac{dy}{dx}+3y^2\frac{dy}{dx}=0$$ $$\frac{dy}{dx}(3y^2-x)=y-3x^2$$ $$\frac{dy}{dx}=\frac{y-3x^2}{3y^2-x}$$
For $P(2,1)$: $$\frac{dy}{dx}=\frac{1-3\times2^2}{3\times1^2-2}=\frac{-11}{1}=-11$$
c) The tangent line to the curve at $P(2,1)$ is $$y-1=-11(x-2)$$ $$y-1=-11x+22$$ $$y=-11x+23$$
The tangent line is also enclosed below.
