Answer
By the proof given
$xy=1$
By the proof given, at points $(2,\frac{1}{2})$ and $(-4,-\frac{1}{4})$ the tangent line are horizontal.
Work Step by Step
Differentiate $x^2y − 2x + 8y = 2$ using chain rule and product rule.
We get, $(x^2)'y+x^2y'-2(x)'+8y'=0$
$\implies 2xy+x^2y'-2+8y'=0$
If the tangent line is horizontal on the curve, then $y'=0$.
$\implies 2xy-2=0$
$\implies xy-1=0$
$\implies xy=1$
Now substitute $y = x^{−1}$ in $x^2y − 2x + 8y = 2$.
We get, $x^2x^{-1}-2x+8x^{-1}=2$
Multiply both sides by $x$.
We get, $x^2x^{-1}x-2x^2+8x^{-1}x=2x$
Now use the exponent product rule.
We get, $x^2-2x^2+8=2x$
Simplify and solve for $x$.
$-x^2+8=2x$
$\implies x^2+2x-8=0$
$\implies x^2+4x-2x-8=0$
$\implies x(x+4)-2(x+4)=0$
$\implies (x+4)(x-2)=0$
$\implies x=-4,2$
So, for $x=-4$ and $x=2$ the tangent line are horizontal.
Now substitute $x=-4$ in $x^2y − 2x + 8y = 2$ and solve for $y$.
$(-4)^2y − 2(-4) + 8y = 2$
$\implies 16y+8+8y=2$
$\implies 24y=-6$
$\implies y=\dfrac{-1}{4}$
Now substitute $x=2$ in $x^2y − 2x + 8y = 2$ and solve for $y$.
$2^2y − 2(2) + 8y = 2$
$\implies 4y-4+8y=2$
$\implies 12y=6$
$\implies y=\dfrac{1}{2}$
Thus, at points $(2,\frac{1}{2})$ and $(-4,-\frac{1}{4})$ the tangent line are horizontal.
