Answer:
x - y = 2
Explanation:
You want an equation for the tangent to (x^2+y^2)^3−8x^2y^2=0 at the point (x, y) = (1, -1).
Inspection
A graph of the curve shows it has a slope of +1 at (x, y) = (1, -1).
In point-slope form the equation of the line is ...
y -k = m(x -h) . . . . . . . . line with slope m through point (h, k)
y -(-1) = 1(x -1) . . . . . . substituting known values
x - y = 2 . . . . . . . . rearranging to standard form
__
Additional comment
Differentiating implicitly, you get ...
3(x^2 +y^2)^2(2x·dx +2y·dy) -16xy^2·dx -16x^2y·dy = 0
at (1, -1), this is ...
3(1 +1)^2(2·dx -2·dy) -16·dx +16·dy = 0
8dx -8dy = 0 . . . . simplified
dy/dx = 1
Then we can proceed with the point-slope equation as above.
<95141404393>