Answer: A
$r = \frac{\partial^2 f}{\partial x^2} = 2y$
$s = \frac{\partial^2 f}{\partial x \partial y} = 2x - 3$
$t = \frac{\partial^2 f}{\partial y^2} = 0$
Since, $rt - s^2 \leq 0,$ (if < 0 then we have no maxima or minima, if = 0 then we can't say anything).
Maxima will exist when $rt - s^2 > 0$ and $r < 0.$
Minima will exist when $rt - s2 > 0$ and $r > 0.$
Since, $rt - s^2$ is never $> 0$ so we have no local extremum.