Hello,
f(x,y)=x²-xy+y²-3=0
If y=0 , x²-3=0 ==>x=-√3 or x=√3
Let's calculate dy/dx
@ means derivate partial.
@f/@x=2x-y
@f/@y=-x+2y
dy/dx=-(@f/@x) /(@f/@y)=-(2x-y)/(-x+5y)
if y=0 and x=√3 then dy/dx=-2√3 / (-√3)=2
idem if y=0 and x=-√3
Equations of the tangent lines
In (√3,0), y-0=2(x-√3)
In (-√3,0), y-0=2(x+√3)