First, we must find the derivative of the curve using implicit differentiation:
![{\displaystyle \displaystyle {\begin{aligned}x^{4}-x^{2}y+y^{4}&=1\\(x^{4}-x^{2}y+y^{4})'&=(1)'\\4x^{3}-(x^{2}y)'+4y^{3}y'&=0\\4x^{3}-x^{2}y'-2xy+4y^{3}y'&=0\\y'(-x^{2}+4y^{3})&=2xy-4x^{3}\\y'&={\frac {2xy-4x^{3}}{4y^{3}-x^{2}}}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/83557947597daa909e3a19ce6b92eaf1e57a001e)
This gives us the equation for the slope. The question is asking us for the slope at the point (-1,1) so we plug in x=-1 and y=1:
![{\displaystyle \displaystyle {\begin{aligned}y'&={\frac {2xy-4x^{3}}{4y^{3}-x^{2}}}\\&={\frac {2(-1)(1)-4(-1)^{3}}{4(1)^{3}-(-1)^{2}}}\\&={\frac {-2+4}{4-1}}\\&={\frac {2}{3}}\\\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/1124c5ef37cf206d1efa23c64aca973a220d0c89)
Therefore, the slope of the tangent line to the curve at (-1,1) is 2/3.