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P( at least one card is a square number)  =

1 - P(neither card is a square number)  =

1 - (4/6) (3/5)  =

1 - 12 /30   =

18 / 30 =   3 / 5

Mary has six cards whose front sides show the numbers  1,2,3,4,5and 6 . She turns the cards face-down, shuffles the cards until their order is random, then pulls the top two cards off the deck. What is the probability that at least one of those two cards shows a square number?

The only square numbers are 1 and 4

P(one of them is 4) = \(\frac{1}{6}\times \frac{5}{5}+\frac{5}{6}\times \frac{1}{5}=\frac{2}{6}\)

P(one of them is 1) = \(\frac{2}{6}\)

P(one is 1 and the other is 4)=\(2*\frac{1}{6}\times\frac{1}{5}=\frac{2}{30}\)

so 

P(at least one is either a 1 or  a 4 is) =  \(\frac{4}{6}-\frac{2}{30}=\frac{18}{30}=\frac{3}{5}\)

Mary has six cards whose front sides show the numbers  1,2,3,4,5and 6 .
She turns the cards face-down, shuffles the cards until their order is random,
then pulls the top two cards off the deck.
What is the probability that at least one of those two cards shows a square number?

\(\color{red}\text{square number}\)

\(\begin{array}{|r|r|r|r|r|r|r|r|} \hline & \text{card 2} & {\color{red}1} & 2 & 3 & {\color{red}4} & 5 & 6 \\ \hline \text{card 1} & & & & & & & \\ \hline {\color{red}1} & & - & \times & \times & \times & \times & \times \\ \hline 2 & & \times & - & & \times & & \\ \hline 3 & & \times & & - & \times & & \\ \hline {\color{red}4} & & \times & \times & \times & - & \times & \times \\ \hline 5 & & \times & & & \times & - & \\ \hline 6 & & \times & & & \times & & - \\ \hline \end{array}\)

\(\text{The probability is $\dfrac{18}{30} = \dfrac{3}{5} \quad (60\ \%) $ } \)

I like CPhill's answer best but they are all correct.