Tens place digit of a perfect square is always even except no ending with $6$ I observed the sequence of perfect squares
$$1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,......$$
I noticed that either unit digit is 6 or tens place digit is even 
Am I right or this doesn't follow for some number
I am trying to find such number or proving it right so please help me!

All numbers can be seen as $10a +b$, where $a\ge 0$ and $0\le b\le 9$. Its square writes $100a^2 + 20 a b + b^2$. The first term has no influence on the last two digits, so you need to consider only $20 a b + b^2$.
Look what happens at the tens place: the term $20 a b$ will necessarily give you an even number, and $b^2$ will give you something. The list of squares of numbers $0$ through $9$ ($1,4,9,16,25,36,49,64,81$) assures you that odd number at tens place comes with $6$ at units place; the final remark is that the term $20ab$ has no impact on the units digit.
