Why is this an illegal start of expression?

Explanation

Your method signature

public static boolean esVocal(Character c)

declares the return type boolean. This means your method is supposed to return something of type boolean.

Solution

If you don't want to return anything you should declare the return type as void:

public static void esVocal(Character c)

Or if you want to return something you need to add the corresponding statement to your method:

public static boolean esVocal(Character c) {
    boolean res = false;

    if (c == 'a' | c == 'o' | c == 'e' | c == 'u') {
        res = true;   
    }

    // You probably wanted to return this
    return res;
}

Notes

Note that you can reduce your method by directly returning the resulting boolean like:

public static boolean esVocal(Character c) {
    return c == 'a' | c == 'o' | c == 'e' | c == 'u';
}

The operator | on numbers is not the logical or. It is a bit-wise inclusive or operation which makes some bit-manipulations on values like:

   0100 1110
or 1000 1011
   ---------
   1100 1111

However on boolean expressions, like in your case, it also performs as logical or (see JLS 15.22.2).

There is also the operator || which always behaves as logical or.

Note that there is a difference between both operators, even when using on boolean values. The | operator will always compute the whole expression, even when the resulting value already is clear. The || aborts evaluation if the result already is clear like in

true || someBoolean || someOtherBoolean

The true already makes the whole result true and || will thus not evaluate the rest of the expression whereas | will.

In your case however you probably want to use ||:

c == 'a' || c == 'o' || c == 'e' || c == 'u'

For the logical and &, && the difference is the same.