Do you mean x²(1/x)? If so lim x→0 is x²/x=x, which →0. Limit is zero.
If you mean (x²)^(1/x), then let f(x)=(x²)^(1/x) or x^(2/x).
f(½)=¼²=1/16=1/2⁴;
f(¼)=(¹⁄₁₆)⁴=1/65536=1/2¹⁶;
f(⅛)=(¹⁄₆₄)⁸=1/2⁴⁸.
As x→0, f(x)→0. Limit is zero.
Put another way:
f(x)=x^(2/x), f(x/2)=(x/2)^(4/x), f(x/4)=(x/4)^(8/x), ...
Eventually, 0<x/2ⁿ<1 for some integer n>0, and f(x/2ⁿ)=(x/2ⁿ)^(2ⁿ⁺¹/x).
Raising a proper fraction to a higher power reduces the fraction, that is, it approaches zero quite rapidly. In this case, the fraction is halved and the power is doubled. We conclude that the limit, therefore, is zero.