All minima should be searched:
f[x_, y_] = x^2 - x + 2 y^2;
g[x_, y_] = x^2 + y^2 - 1;
First we look for the minima on the edge. With Langrange we get:
L = f[x, y] + λ*g[x, y];
sol = Solve[{Grad[L, {x, y}] == 0, g[x, y] == 0}, {x, y}, λ, Reals]

hessian = D[f[x, y], {{x, y}, 2}] /. sol;
hessian: positive definite -> strict local minimum
hessian: negative definite -> strict local maximum
test = Transpose[{{x, y, f[x, y]} /. sol,
PositiveDefiniteMatrixQ /@ hessian,
NegativeDefiniteMatrixQ /@ hessian}];
TableForm[Partition[Flatten[test], 5], TableAlignments -> Right,
TableHeadings -> {None, {"x", "y", "f[x,y]", "pos. definite", "neg. definite"}}]

All points are minima!
Now we are looking for minima within the region.
rsol = First@Solve[Grad[f[x, y], {x, y}] == 0, {x, y} \[Element] ImplicitRegion[x^2 + y^2 < 1, {x, y}]]

rhessian = D[f[x, y], {{x, y}, 2}] /. rsol;
{PositiveDefiniteMatrixQ @ rhessian, NegativeDefiniteMatrixQ @ rhessian}
{True, False}
{f[x, y] /. rsol , rsol}

This is the global minimum.
FindMinimum
usually does what is required. What do you mean by exact? $\endgroup$ – Hugh Apr 22 '20 at 10:27Reduce[{D[x^2 - x + 2*y^2 D[x^2 - x + 2*y^2, x] == 0, x] == 0, D[x^2 - x + 2*y^2, y] == 0, x^2 + y^2 <= 1}, {x, y}, Reals]
. Its behavior on the boundary should be studied separately. $\endgroup$ – user64494 Apr 22 '20 at 10:53FindMinimum
might also find a global min and miss local ones. To get all of them in some region one can set up a KKT solver. $\endgroup$ – Daniel Lichtblau Apr 22 '20 at 13:07