Is this language $\{0^n1^m | (n+m) \text{is even}\}$ regular or not?
You seem to have misunderstood the pumping lemma.

Taking the definition from Introduction to the Theory of Computation by Michael Sipser


  If $A$ is a regular language, then there is a number $p$ (the
  pumping length) where if $s$ is any string in $A$ of length at least $p$, then $s$ may be divided into three pieces, $s = xyz$, satisfying the following conditions:
  
  
  for each $i ≥ 0, xy^i z \in A$,
  $|y| &gt; 0$, and
  $|xy| \le p$.
  


A few points:


The lemma only says that a pumping length $p$ exists. You do not get to choose the value of $p$. So, you cannot say that let $p=3$.
For every string $s$ in the language with length at least $p$, it may be divided into three pieces, $s=xyz$. Notice that this again talks only of the existence of a way to split $s$ into $xyz$ given that $A$ is regular. So, you do not get to choose how the string splits into $x$, $y$, and $z$ in the general case. Finding a split that does not satisfy the conditions does not show that the language is not regular. You need to find such a string that cannot be split as stipulated by the definition to show that a language is not regular.
Finally, the pumping lemma does not say what happens if $A$ is not regular. So, even a non-regular language may satisfy it, and the pumping lemma cannot be used to show that a language is regular.