

Setting both partial derivatives to 0 gives a single critical point at

, which does fall inside the unit disk.
At this point, the value of the derivative of the Hessian matrix is

while the value of the second-order partial derivative with respect to

is

This means the critical point is the site of a local minimum, so this is the coldest point on the plate with a temperature of

.
The hottest point on the plate must then be found on the boundary. Let

and

, so that


Then the boundary of the plate (the circle

) is a function of a single variable

considered over

. Differentiating once gives


You'll find that

attains three extrema on the interval

, with relative maxima at

and

and a relative minimum at

(and

, if you want to include that).
We already found our minimum on the inside of our plate - which you can verify to have a lower temperature than at the points given by

- and we find two maxima at

and

, each giving a maximum temperature of

.
Converting back to Cartesian coordinates, these points correspond to the points

.