I want to show that $f(x,y)=x \log(y),\,x\in[0,1],y\in(0,1]$ is concave so $-f(x,y)$ should be convex. I write the Hessian matrix as follows $$\mathcal{H}=\begin{bmatrix}-\frac{\partial^2 f}{\partial x^2} & -\frac{\partial^2 f}{\partial x\partial y}\\ -\frac{\partial^2 f}{\partial x\partial y} & -\frac{\partial^2 f}{\partial y^2} \end{bmatrix}=\begin{bmatrix} 0 & -\frac{1}{y}\\ -\frac{1}{y} & \frac{x}{y^2} \end{bmatrix}$$ Then $\mathcal{H}$ should be positive semi-definite. Necessary and sufficient condition is that all of its principal minors should be non-negative including its determinant. However $$|\mathcal{H}|=0-\frac{1}{y^2}=-\frac{1}{y^2}\leq 0$$ But when I plot $f(x,y)$ it is concave in the range $x\in[0,1],y\in(0,1]$. I don't know where I am wrong.
Note that the minor consisting of just the lower-right entry is positive, but as you noted the full determinant is negative. Consequently, the function is neither convex nor concave.
To be more concrete about that last assertion, consider the function along the paths $y=x$ and $y = 1-x$. The corresponding contours on $f$ are $f(x,x) = x \log(x)$ and $f(x, 1-x) = x \log(1-x)$, which lie entirely above and below their respective secant lines. Thus, the function is indeed neither convex nor concave.
In final full disclosure, I cannot easily see the facts I just cited from the graph.