Yes, there are. I am trying to explain step by step.
Say we want the square of n = 67. First of all, the square of 7 is 49. Write 9 in the answer place and keep 4 for carrying over process.
∴n^2 = 9
Then, see that 6 and 7 form the number 67. Their product is 42. The rule says you have to double it. Hence you get double of 42 as 84. Also, you have 4 in your hand for carrying over. Together you get 84+4=88. Write the unit digit eight beside 9 and keep 10s digit 8 for next round carry-over process.
∴ n^2 = 89
Finally, the 10s digit is 6. Square it, and you will get 36. Plus you have 8 in your hand. Together we get 36+8, i.e. 44. Now write it beside 89, and we are done.
∴ 67^2 = 4489.
Try your self: 39^2, 23^2, 52^2.
 
Verify your answer using a calculator.
 
For squaring three digits or more, there is procedure indeed, but I can't explain them here within short place. 
 
Although I can show you some other special technique for squaring 2 digit numbers of special types. 
 
1. If the number ends with 5: 
Let's say we need to square 85. First, write 25 straight forward in answer place. 
85^2 = 25
 
Then, note that after ignoring five we are left with the number 8 in the given number 85. The next number of 8 is 9. The rule says to make the product which is in our present case is 8 x 9, i.e. 72.
 
Write it down, and we are done.
85^2 = 7225.
Similarly, 95^2 = 9025 because 9 x 10 = 90 and 25 already is written etc.
115^2 = 13225 because 11 x 12 = 132. Etc
 
2. If the number is near to 100.
Let's say we need to square 93. 
 
The difference between 93 and 100 is 7. The rule says to write down the square of 7 first, which is here 49.
Next, decrease 93 by 7. Which gives 86. Hence the final answer becomes 
93^2=8649.
Similarly, 95^2 is 9025 because the difference between 95 and 100 is 5. The square of 5 is 25, and if we decrease 95 by 5, we shall get 90
 Together we thus obtain 
95^2= 9025.
Notice we have achieved the same already before by using the general squaring technique explained above.
 
Caution. Suppose we want to square 98. This time doesn't be tempted to write 964. That's not correct. While you are squaring the difference between 100 and the given number, make sure the answer occupies exactly two places only. If not, use 0 to fill up the lattice. Hence 98^2 is not 964 but in 9604.
 
Similarly, for 88^2, first, write down the square of the difference which is 12^2, i.e. 144. But this time it is a three-digit answer. So keep the last two digits and carry over the excess digit which is 1 in this present case. Now decrease 88 by 12, and we get 76. Added by carried over one we get 77. Hence 88^2 = 7744.
 
There are more so many special rules. But we can discuss them some other days.
 
Take care.

Yes. Every number is special.

you can use algebraic identity to find to the square of any number.

First we see about the number end with 9..

Eg:1) 99=100-1

(100-1)^2=100^2-2(100)(1)+1^2

                =10000-200+1

                =9800+1

                =9801.

Eg:2) 29=30-1

(30-1)^2=30^2-2(30)(2)+1^2

               =900-120+1

               =780+1

               =781

If you practice this directly, start from 2nd step.

This method uses the identity (a + b) ² = a ² + 2ab + b ² 
Step 1 : Find 57² 
Here a = 5 and b =7 

Step II: Underline the digit of b ² ( in column III) and add its tens digit, if any, to 2 x a x b (in column III)

Step III: Underline the digit in column II and add the number formed by tens and other digit, if any, to a ² in column I. 

Step IV: under the number in column I 

Write the underlined digits from the unit digit.
Therefore, 57 ² = 3,249 .