Assume [tex]y=y(x)[/tex].
[tex]7x^2+xy+7y^2=15[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dx}[7x^2+xy+7y^2]=\dfrac{\mathrm d}{\mathrm dx}[15][/tex]
[tex]14x+y+x\dfrac{\mathrm dy}{\mathrm dx}+14y\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}(x+14y)=-(14x+y)[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{14x+y}{x+14y}[/tex]
The slope of the tangent line to the curve at [tex](a,b)[/tex] is then [tex]-\dfrac{14a+b}{a+14b}[/tex], so the tangent to [tex](1,1)[/tex] has slope
[tex]\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=-\dfrac{15}{15}=-1[/tex]
The point-slope form of the tangent line is then
[tex]y-1=-(x-1)\iff y=-x+2[/tex]
