Symmetry

$(x^3+1-xy^2-y^2)-(y^3+1-yx^2-x^2)=0\\x^3-y^3-xy^2+yx^2-y^2+x^2=0$ Factorizing,we get $(x-y)(x+y)(x+y+1)=0$ .Now we have 3 cases: $\text{Case 1:} x=y\\ \implies (y)^3+1-(y)(y^2)-y^2=0\\ \implies y^2=1\implies y=1 \; \text{or}\; y=-1$ $y=1$ gives $(x,y)=(1,1)$ while $y=-1$ gives $(x,y)=(-1,-1)$ $\text{Case 2:} x=-y\\ \implies (-y)^3+1-(-y)y^2-y^2=0\\ \implies y^2=1\implies y=1\;\text{or}\; y=-1$ $y=1$ gives $(x,y)=(-1,1)$ while $y=-1$ gives $(x,y)=(1,-1)$ $\text{Case 3:} x=-1-y\\ \implies (-1-y)^3+1-(-1-y)y^2-y^2=0\\ \implies -3y^2-3y=0\implies -3y(y+1)=0$ The case $y=0$ gives $(x,y)=(-1,0)$ while the case $y=-1$ gives $(x,y)=(0,-1)$ .

All the 6 pairs we have found satisfy both equations.Hence the answer is $\boxed{6}$ .