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I am working on some basic field question for a linear algebra class.Below is the question. I looked at the multiplication of the $\mathbb Z_7$ field, and I don't see any of the elements whose square is $3$ . Is this the correct way to look at it?

Let $p$ be a prime integer and $a\in\mathbb F_p$. Does there necessarily exist $b\in\mathbb F_p$ satisfying $b^2=a$.

Thanks in advance for any insight.

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    This may be of interest to you: https://en.wikipedia.org/wiki/Legendre_symbol – 2017-01-29
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    In your title you probably mean «Existance of a $b$ such that...» As it is written, it does not make sense. – 2017-01-29
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    This is a good way to look at it. You can even find a counterexample with $p=3$. – 2017-01-29

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3 has no square root in $\mathbb{F}_7$. You may use the Lagrange symbol to check this but it seems overkill to me. You can check it directly by evaluating all squares of $\mathbb{F}_7$.

Not every element of a finite field $\mathbb{F}_p$ has a square root, for any $p>2$. You can see it by a combinatorial argument: For each nonzero $a$, $a\neq -a$ and $a^2=(-a)^2$. Moreover, $a^2=b^2$ implies $a=\pm b$. Now we can conclude that the map $x\mapsto x^2$ is two-to-one map over $\mathbb{F}_p^\times$. Especially there are $(p-1)/2$ nonzero elements which has no square root.

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Let $p$ be a prime integer and $a\in\mathbb F_p$. Does there necessarily exist $b\in\mathbb F_p$ satisfying $b^2=a$.

The answer is no, and you've already found a counterexample. Take $p=7$ and $a=3$. \begin{align*} 0^2 &= 0\\ 1^2 &= 1\\ 2^2 &= 4\\ 3^2 &= 9 \equiv 2 \pmod 7\\ 4^2 &= 16 \equiv 2 \pmod 7\\ 5^2 &= 25 \equiv 4 \pmod 7\\ 6^2 &= 36 \equiv 1 \pmod 7 \end{align*} So there is no $b \in \Bbb F_7$ such that $b^2 = 3$. There may be simpler counterexamples but one is all you need.

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Let $\mathbb{F}$ be a finite field with characteristic $\neq2$ (so that $-1\neq1$), and consider the function $s:\mathbb{F}\to\mathbb{F}$ given by $s(b):=b^2$. As $s(-1)=s(1)=1$ it follows that $s$ cannot be injective, which (as the domain and codomain of $s$ have the same finite size) implies that $s$ cannot be surjective; i.e. there is some $a\in\mathbb{F}$ such that $s(b)=b^2\neq a$ for every $b\in\mathbb{F}$.