For what pairs $a,b\in\mathbb{\mathbb{C}}$ is the line $L(x,y)=ax+by+1=0$ tangent to the curve $C(x,y)=x^4+y^4+1=0$?
By definition of "tangent", if I have a point $(c,d)\in C$, and a line parametrized by $(c+pt,d+qt)$, then tangency is when $C(c+pt,d+qt)$ has $t=0$ as a root of multiplicity at least $2$.
Here the line is given in the form $ax+by+1=0$, which is harder to parametrize, and it's also hard to find the intersection points between $L(x,y)$ and $C(x,y)$. How to proceed?
I don't think the line is that hard to parametrize that you should give up on that approach. But anyway, here is a different one.
Two manifolds are tangent if their tangent spaces coincide (in case of unequal dimensions, are related by inclusion). Equivalently, one can consider the orthogonal complement of the tangent space, which in the present case is the line given by the gradient of the implicit equation: $\langle a,b\rangle$ for $L$ and $\langle 4x^3,4y^3\rangle$ for $C$. For these two to be parallel, we need $a=\lambda x^3 $ and $b=\lambda y^3 $ for some $\lambda\in\mathbb C$. From the equation of $L$, $$\lambda x^4+\lambda y^4 + 1 =0$$ which in view of the equation of $C$ gives $\lambda=1$.
So, $(a,b) = (x^3,y^3)$ where $x,y$ are subject to $x^4+y^4+1=0$. In other words, the condition on $a,b$ is that $a^{4/3}+b^{4/3}+1$ is zero for some choices of these cubic roots.