Let $\beta\in\mathbb{R}^{d}$, we consider the function $T:\mathbb{R}^{d}\times \mathbb{R}\rightarrow \mathbb{R}$ given by $$T(x,y):=\left\langle \theta , (x,y) \right\rangle - (y-\beta^{t}x)^{2}$$
where $\theta\in\mathbb{R}^{d+1}$ is fixed. (we recall $\left\langle \theta , v \right\rangle=\theta^{t}v$)
Find the derivative of $T$ and determine for which values of $(x, y)$ we have $T'(x,y)=0$.
My attempt: If we consider $(h,g)\in \mathbb{R}^{d}\times \mathbb{R}$, we know that $$T((x,y)+(h,g))=T(x,y)+T'(x,y)(h,g)+R(h,g)$$ where $\frac{R(h,g)}{\left\|(h,g)\right\|}\rightarrow 0$ when $\left\|(h,g)\right\|\rightarrow 0$. In this sense, we have $$T((x,y)+(h,g))=\left\langle \theta , (x,y) \right\rangle - (y-\beta^{t}x)^{2}+\left\langle \theta , (h,g) \right\rangle-2(y-\beta^{t}x)(g-\beta^{t}h)-(g-\beta^{t}h)^{2}.$$ I am not sure of the choice of $T'$ and $ R $. For me $$\begin{array}{rcl}T'(x,y):\mathbb{R}^{d}\times\mathbb{R} &\rightarrow&\mathbb{R}\\ (h,g)&\mapsto& \left\langle \theta , (h,g) \right\rangle-2(y-\beta^{t}x)(g-\beta^{t}h). \end{array} $$
My problem is determine for which values of $(x, y)$ we have $T'(x,y)=0$.
The motivation:
My motivation to address this problem is to determine for which values of $\theta$ we have $$\sup_{(x,y)\in \mathbb{R}^{d}\times \mathbb{R}}\left[\left\langle \theta , (x,y) \right\rangle - (y-\beta^{t}x)^{2}\right]<\infty \tag{1}$$ My strategy is to find this supremum in the traditional way, deriving and equaling to zero, then finding the values where that supremum is achieved and replacing them in $\left\langle \theta , (x,y) \right\rangle - (y-\beta^{t}x)^{2}$ to know the value of the supreme (note that $\left\langle \theta , (x,y) \right\rangle - (y-\beta^{t}x)^{2}$ is concave respect to $(x,y)$).
I do not know if there is an easier way to do all this. I feel that the fact of being $\left\langle \theta , (x,y) \right\rangle - (y-\beta^{t}x)^{2}$ concave respect to $(x,y)$ we have (1) for any $ \theta\in\mathbb{R}^{d+1}$, but I'm not sure.