Answer
point $(1,1)$, slope is $-\frac{1}{2}$
point $(1,-1)$, slope is undefined.
Work Step by Step
Step 1. Given $x^3y^3+y^2=x+y$, we have $3x^2y^3+3x^3y^2y'+2yy'=1+y'$ or $y'=\frac{3x^2y^3-1}{1-2y-3x^3y^2}$ which gives the slope of tangent lines to the curve.
Step 2. At point $(1,1)$, we can easy check to see that the point is on the curve and the slope of the tangent line is $m_1=\frac{3(1)^2(1)^3-1}{1-2(1)-3(1)^3(1)^2}=-\frac{1}{2}$ and the equation is $y-1=-\frac{1}{2}(x-1)$ or $y=-\frac{1}{2}x+\frac{3}{2}$
Step 3. At point $(1,-1)$, we can easy check to see that the point is on the curve and the slope of the tangent line is $m_2=\frac{3(1)^2(-1)^3-1}{1-2(-1)-3(1)^3(-1)^2}=-\frac{4}{3-3}=\infty$ and this slope is undefined.
