Answer
$(y-1)=\frac{3}{4}(x-2)$
Work Step by Step
$(x^2-xy-y^2=1)'$
Using the chain rule:
$2x-xy'-y-2yy'=0$
When we solve for $y'$, we get:
$y'=(\frac{2x-y}{2y+x})$
Plug in the given point $(2, 1)$ to get the slope of our tangent line.
$y'=\frac{3}{4}=m$
Using point-slope form:
$(y-1)=\frac{3}{4}(x-2)$
