Find a normal from an implicit derivative

The complete working appears in both copies. Implicit differentiation and the product rule give $3x^2+3y^2\dfrac{dy}{dx}+3x\dfrac{dy}{dx}+3y=0$. Thus $(3y^2+3x)\dfrac{dy}{dx}=-3x^2-3y$, so $\dfrac{dy}{dx}=-\dfrac{x^2+y}{x+y^2}$ where the denominator is nonzero.

At $(2,-1)$ the denominator is $2+(-1)^2=3\ne0$, and $dy/dx=-(4-1)/3=-1$. This is the tangent slope, not the normal slope.

The normal slope is $-1/(-1)=1$. Through $(2,-1)$ its equation is $y-(-1)=1(x-2)$, giving $y=x-3$, as printed.