There is no obvious way to apply Eisenstein's criterion; and if I assume by contradiction that $x^6-xy+y^6=f(x,y)g(x,y)$, f with homogeneous degree $\leq 3$. Then I have that $f(x,y)=\sum_{i,j} a_{ij}x^i y^j$, $g(x,y)=\sum_{i,j} b_{ij}x^i y^j$, so $$f(x,y)g(x,y)=\sum_{i,j,k,l}a_{ij} b_{kl} x^{i+k} y^{j+l}=\sum_{i,j} x^i y^j\left(\sum_{i_0+i_1=i}\sum_{j_0+j_1=j}a_{i_0 j_0} b_{i_1 j_1} \right) =x^6-xy+y^6$$ so that $$\sum_{i_0+i_1=i}\sum_{j_0+j_1=j}a_{i_0 j_0} b_{i_1 j_1}= \left\{\begin{array}{ll} 1 & \quad i=6,j=0\\ -1 & \quad i=j=1\\ 1 & \quad i=0,j=6\\ 0 & \quad \mathrm{else} \end{array}\right.$$
which doesn't seem like a particularly fruitful approach. Other than painstakingly writing out the 10 possible terms for $f$ and the 28 possible terms for $g$ to get a horrendous 38-variable quadratic system of equations, I was wondering if there was some easier way to do this.
