What is definition of independent random variable
No, NO and $\mathbf{NO}$.

Knowing that $E[X]=E[Y]=0$ and $E[XY]=0$ (or more generally that $E[XY]=E[X]E[Y]$ or equivalently, $E[XY]-E[X]E[Y]=0$) does not help in the least in proving or deducing that $X$ and $Y$ are independent random variables.  The conditions stated in the previous sentence are necessary for $X$ and $Y$ to be deemed independent random variables, but they are a very far cry from being sufficient for independence to hold. That is, if $E[XY]=E[X]E[Y]$, then we might suspect that $X$ and $Y$ might be independent random variables, but we cannot jump to the conclusion that they are independent random variables based on just the evidence that $E[XY]=E[X]E[Y]$.


  Definition: $X$ and $Y$ are said to be independent random variables (sometimes more prolixly mutually independent random variables) if their joint distribution function (joint CDF) $F_{X,Y}(u,v) = P\{X \leq u, Y\leq v\}$ satisfies
  $$F_{X,Y}(u,v)=F_{X}(u)F_{Y}(v) ~\text{for all real numbers}~ u ~\text{and}~v.\tag{1}$$
  Here, $F_{X}(u)=P\{X \leq u\}$ and $F_{Y}(v)=P\{Y\leq v\}$ are the marginal CDFs of $X$ and $Y$ respectively. It is important to remember the qualifier to the equality $F_{X,Y}(u,v)=F_{X}(u)F_{Y}(v)$: it must hold at all points in the $u$-$v$ plane.. Inanities such as "$X$ and $Y$ are independent when $X=5$ and $Y=3$ but not otherwise" are not permitted.


If $X$ and $Y$ are discrete random variables, then $(1)$ is equivalent to their joint _probability mass function (joint pmf) $p_{X,Y}(u_i, v_j)$ factoring into the product $p_X(u_i)p_Y(v_j)$ of the marginal pdfs
for all $i$ and $j$. 

If $X$ and $Y$ are jointly continuous random variables, then $(1)$ is equivalent to their joint _probability density function (joint pdf) $f_{X,Y}(u, v)$ factoring into the product $f_X(u)f_Y(v)$ of the marginal pdfs
_for all real numbers $u$ and $v$. 

Now, as the OP correctly points out, if $X$ and $Y$ are independent, then it is true that $E[XY]=E[X]E[Y]$, that is,
$$X, Y ~\text{independent random variables} \implies E[XY]=E[X]E[Y]$$ but 


  the reverse implication
  $$E[XY]=E[X]E[Y]\implies X, Y ~\text{independent random variables}$$
  $$\mathbf{DOES~NOT~HOLD}$$


Still don't believe a word of all this malarkey? Consider discrete random variables $X$ and $Y$ such that 
$$p_{X,Y}(1,0) = p_{X,Y}(-1,0) = p_{X,Y}(0,1) = p_{X,Y}(0,-1) = \frac 14.$$
Verify that 


$X$ and $Y$ take on values $\pm 1$ with equal probability $\frac 14$ and value $0$ with probability $\frac 12$.
$XY=0$ always and so $E[XY]=0$ also, and that $E[X]=E[Y]=0$ also.
The joint pmf does not factor into the product of the marginal pdfs for at any of the four points where the joint pmf is nonzero.