Question: We have $k\_1:= \mathbb F\_7(\alpha)$ and $k\_2 := \mathbb F\_7(\beta)$ where $\alpha^2 = 3$ and $\beta^2 = -1$ in $\mathbb F\_7$. I have to show that these two are isomorphic. 

Let $\phi:k\_1 \rightarrow k\_2$ be a homomorphism which preserves $1 \in k\_1$. Then 

$$\phi(\alpha^2)= \phi(3) = 3 = \phi(\alpha)^2$$

where 

$$\phi(\alpha) = x + y\beta\;,\;\;x,y \in \mathbb F\_7$$

Thus 

$$(x+y\beta)^2 = \phi(\alpha)^2 = x^2 + 2xy\beta +y ^2 \beta^2 = x^2 +2xy\beta -y^2=3$$

So $x$ or $y$ must be $0$. But $y$ can't be zero because $3$ has no root in $\mathbb F\_7$. So $x = 0$ such that 

$$-y^2 = 3 \rightarrow y \in \{-2, 2\}$$

Is the function $\phi$ with $\phi(x) = x\;$ for $\;x \in \mathbb F\_7\;$ and $\;\phi(\alpha) = 2\beta\;$ then an isomorphism ?

Answer: 
Well, you have a candidate; why don't you check whether $\phi((a+b\alpha)(c+d\alpha)) = \phi(a+b\alpha)\phi(c+d\alpha)$, for all $a,b,c,d \in \mathbb{F}\_7$? For sums it's trivial, as you already define it as a linear map over the base field.
An idea: since

$$k\_1=\Bbb F\_7[x]/\langle\;x^2-3\;\rangle\;,\;\;k\_2=\Bbb F\_7[x]/\langle\;x^2+1\;\rangle$$

Now, in both cases there exists a *primitive element*, i.e.: an element that generates the multiplicative group $\,k\_i^\*\;$ , so if you find both elements in both fields you shall be done...

For example: let $\,\alpha\in\overline{\Bbb F\_7}\;$ be a root of $\,x^2-3\in\Bbb F\_7[x]\,$ , then in $\,k\_1\,$ we have
$$\alpha^2=3\implies k\_1:=\{p(\alpha)\;;\;p(x)\in\Bbb F\_7[x]\;,\;\;\deg p\le 1\}$$

For example, we have that $\,\alpha +5\;,\;3\alpha\;,\;2\alpha -5\,$ , etc. are element in $\,k\_1\,$ . 

Since $\,|k\_1^\*|=48\;$ ,every non-zero element in the field has *multiplicative* order a divisor of $\,48\,$ , so we can try the following: pick (nd remember: we work modulo $\,7\,$ all the time!)

$$\begin{align\*}r:&=\alpha+1\\
 r^2&=\alpha^2+2\alpha+1=3+2\alpha+1=2\alpha+4\\
r^4&=4\alpha^2+2\alpha+2=2\alpha\\
r^6&=(2\alpha)^3=\alpha(3)=3\alpha\\
r^{12}&=(3\alpha)^2=2\cdot 3=-1\end{align\*}$$

and we've found that $\;r\;$ is **not** a generator of $\,k\_1^\*\,$ (why? check what happens with $\,r^{24}\,$ ...) . Try other elements until you find an generator and then try to do the same with $\,k\_2\,$ and then you'll have an easy choice for your isomorphism...