= Solution
Use the <Lagrangian>
$$
\mathcal L=x^4+4y^4-\lambda(x^2-y^2-1).
$$
The <Lagrange multiplier> equations are
$$
4x^3-2\lambda x=0,\qquad
16y^3+2\lambda y=0,\qquad x^2-y^2=1.
$$
Feasibility makes $x\ne0$, so $\lambda=2x^2>0$. The second equation becomes $2y(8y^2+\lambda)=0$, forcing $y=0$ and then $x=\pm1$. The constraint <gradient> $(2x,-2y)$ is nonzero on the feasible set, so the multiplier calculation has no missing singular feasible point.
To verify the global minimum rather than just stationarity, eliminate $x^2=1+y^2$:
$$
x^4+4y^4=1+2y^2+5y^4\ge1.
$$
Equality holds exactly at \b[$(x,y)=(1,0)$ or $(-1,0)$, with minimum value $1$].
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