How $x^4$ is strictly convex function?

If the second derivative is strictly positive, then the function is strictly convex. However, the converse need not be true. A function $f:\Bbb R\to\Bbb R$ is strictly convex if and only if for all $x,y\in\Bbb R$ with $x\neq y$ we have $$f\bigl(tx+(1-t)y\bigr)<tf(x)+(1-t)f(y)$$ for all $0<t<1$. $f(x)=x^4$ is indeed strictly convex.

This is a common misconception. Many make the same mistake regarding the relationship between positive first derivative and increasing functions (the former implies the latter, but not vice versa). See here and here, for examples of people making such errors.

Strict Convexity is when $f(tx+(1-t)y)<tf(x)+(1-t)f(y)$. The fact that $f''(x)\geq 0$ implies $f$ is convex, however, it does not necessarily imply that $f$ is not strictly convex. In the case of $f(x)=x^4$, you get $f''(x)=12x^2\geq 0$ but in fact $f$ is strictly convex. Indeed, any line segement lies above the curve.