Is A XNOR B logically equivalent to A XOR (NOT B)?
A XNOR B:
$$ \overline{A \oplus B} = \overline {A\cdot \overline B +  \overline A \cdot  B}$$  

A XOR ~B:
$$ A \oplus \overline B = A\cdot B +  \overline A \cdot  \overline B$$

Starting from A XNOR B, and using De Morgan's law:
$$ \overline{A \oplus B} = \overline {A\cdot \overline B +  \overline A \cdot  B} \overset{DM}{=} {(\overline A + B) \cdot ( A + \overline B)} = { ( (\overline A + B) \cdot A + (\overline A + B) \cdot \overline B)}$$
$$ \overline{A \oplus B} = { ( (\overline A \cdot A+ B \cdot A)  + (\overline A  \cdot \overline B+ B \cdot \overline B))} =  (\overline A \cdot A+ B \cdot A)  + (\overline A  \cdot \overline B+ B \cdot \overline B) $$

Notice some of those terms are contradictions, like \$\overline A \cdot A=0\$,se we can replace them
$$ \overline{A \oplus B} = 0 +B \cdot A  + \overline A  \cdot \overline B +0.$$

With similar manipulations you should be able to get to the final answer, and even show the reverse (even though the steps are bidirectional).