I have the following two-variable function $h(t,w):= t\,\log\left(\dfrac{w}{t}\right), \, t\in\mathbb{R}^{>0},\, w\in\mathbb{R}^{>0} $ and I was trying to prove whether this function is strictly concave or not. To check the strict concavity, one can form find the related Hessian matrix and check to see whether the Hessian matrix is negative definite or not.
The Hessian matrix, denoted by $H$, related to $h(t,w)$ is given by \begin{equation} H\stackrel{\triangle}{=} \begin{bmatrix} \dfrac{-1}{t} & \dfrac{1}{w}\\ \dfrac{1}{w} & \dfrac{-t}{w^2} \end{bmatrix}, \end{equation}
To check the negative definiteness of $H$, I chose an arbitrary two-dimensional column vector $\mathbf{x} = [x_1,x_2]^T \in\mathbb{R}^2$ where $\mathbf{x} \neq \mathbf{0}$, and found that
\begin{equation} \mathbf{x}^T A \mathbf{x} = -\frac{1}{t}\left[\left(x_1 - x_2 \frac{t}{w}\right)^2\right]\leq 0. \end{equation} We clearly observe that $\mathbf{x}^T A \mathbf{x}\leq 0$ since $t>0,\, w>0 $. Also the equality holds whenever $\frac{t}{w} = \frac{x_1}{x_2}$. Does this imply that equality happens only when the ratio $\frac{t}{w}$ is fixed? The equality presents a dependency between the arbitrary chosen vector $\mathbf{x}$ and the arguments of the functions. In this case, can we declare whether this function is strictly concave?
Usually you cannot say anything about strict concacvity if you have a negative semi-definite (as opposed to negative definite) Hessian. But in this case, your analysis gives us a very simple counterexample: restrict your function to the ray $\{(t,t):t>0\}$, where the function becomes $f(t,t)=0$, which is constant over a line, and hence cannot be strictly concave.
To get a precise counterexample, take any three points on the line (say $(1,1),(2,2),(3,3)$) and use the definition of strict concavity.
PS: You could have made the same inference by restricting to any ray, not only the one I used.