There's one eye in the corner already, so the problem is basically how to destroy the potential eye in the middle.
is the key point here. One defense would be 'a' and 'b', as Black can play 'a' next to capture the three stones.
After , there are two potential eyes at around 'a' and 'b'. Since 'c' and 'd' are miai to destroy the eye at 'a', we start with the one on the right.
After , 'a' and 'b' are miai to destroy the eye on the right while 'c' and 'd' are miai to destroy the eye on the left.
If , is nice. White can't make a second eye. If White 'a', Black throws in at 'b'.
Black must play the two throw-ins in the correct order. If black tries the other order, then white doesn't need to capture the second stone:
at a would be a mistake, transposing back to the solution. But this lives, with half eyes at both and .