How to solve this inequality , when there is a irrational power?

You don't have to worry about $x < 3$ because it is not in the domain. Irrational powers are not defined when the base is negative.

You might remember that for an irrational number $a$, by definition the exponentiation function with exponent $a$ is defined by $$t^a = e^{a \ln(t)} $$ and the domain of that function is the positive axis $t > 0$.

So for your function, the domain is $x > 3$.

Hint: For $x<3$, $(x-3)^{\sqrt 2}$ is not defined, so you must impose the condition $x\ge 3$ from the start. Aftwerwards you may proceed as you did.