Both the first and third terms are perfect squares.
$$ 36x^2 = \left( \color{blue}{ 6y } \right)^2 ~~ \text{and} ~~ 1 = \left( \color{red}{ 1 } \right)^2 $$The middle term ( $ 12x $ ) is two times the product of the terms that are squared.
$$ 12x = 2 \cdot \color{blue}{6y} \cdot \color{red}{1} $$We can conclude that the polynomial $ 36y^{2}+12y+1 $ is a perfect square trinomial, so we will use the formula below.
$$ A^2 + 2AB + B^2 = (A + B)^2 $$In this example we have $ \color{blue}{ A = 6y } $ and $ \color{red}{ B = 1 } $ so,
$$ 36y^{2}+12y+1 = ( \color{blue}{ 6y } + \color{red}{ 1 } )^2 $$