Consider a problem $X$ can be reduced to 3-partition problem. So, when 3-partition has a solution then $X$ has a solution. But if 3-partition does not have a solution, they $X$ may or may not have a solution. In this case, is it legal to say that the problem $X$ is np-complete?
No, that doesn't prove NP-completeness.
The thing you describe isn't a reduction. It must be that the instance of $X$ has a solution if, and only if, the instance of 3-partition has a solution.
Reducing $X$ to 3-partition just proves that $X$ is in NP. To prove completeness, you need to reduce an NP-complete problem to $X$.
I suggest that you revise the basic concepts from your class notes, textbook or online resources.