Use implicit differentiation to find an equation of the tangent line to the curve at the given point 
Use implicit differentiation to find an equation of the tangent line to the curve $$x^2+xy+y^2=1$$ at  $(1,1)$.

I am not sure how I should work this out because the given point is not on the curve.

Given $\ x^2 + xy + y^2 = 1$, find tangent line thru (1,1)
$\ Solve \ for \ y $
$ y = \frac{-x \pm \sqrt {4-3x^2}}{2}= \frac{-x}{2}\pm\frac{\sqrt{4-3x^2}}{2}=f(x)$
$\frac{dy}{dx}=\frac{-1}{2}\mp\frac{1.5x}{\sqrt{4-3x^2}}=f'(x)$
Let point (a, f(a)) is tangent point; for the slope, we have the equation
$\ f'(a)=\frac{f(a)-1}{a-1}$
Solve this for a , we obtain
$ a \approx -0.483 \ ; f'(a) \approx -0.10118$
The tangent line y = f '(a)(x - 1) + 1
$y = -0.10118(x - 1) + 1$
By inverse symmetry, also $y = -0.483(x - 1) + 1$
desmos pic
