What does the unit sphere for the norm on $\mathbb{R}^2$, $\displaystyle N(x,y)\rightarrow\sup_{t\in\mathbb{R}}\frac{|x+ty|}{t^2+t+1}$, look like ?
My approach was to consider $y=ax$ so as to get $f(y)=|x|\dfrac{|1+aw|}{t^2+t+1}$ then study the variations of this function wrt $t$ and hence deduce the points on the line $y=ax$ where the norm is $1$. By doing this for all lines that would have given me the whole unit ball. However, with this method, I encounter a determinant $5a^2-2a+1$ which would require branching many cases, hence I don't think that's the way to go.
We have
$$(x,y)\in B(0,1)\iff N(x,y)=\sup_{t\in\mathbb{R}}\frac{|x+ty|}{t^2+t+1}\le1\iff |x+ty|\le t^2+t+1, \forall t\in\Bbb R\\\iff -t^2-t-1\le x+ty\le t^2+t+1,\forall t$$
the second inequality gives
$$t^2+(1-y)t+1-x\ge0,\forall t\iff \Delta_2=(1-y)^2+4x-4\le0$$ and the first inequality give $$t^2+(1+y)t+x+1\ge0,\forall t\iff \Delta_1=(1+y)^2-4x-4\le0$$ hence the unit ball is the set of point $(x,y)$ satisfying $$\Delta_1\le0\quad;\quad \Delta_2\le0$$
We plot the unit ball using Walfram alpha