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.
Hint: $ $ mod $\,7\!:\ \alpha^2 \equiv 3 \equiv -4 = 4\beta^2\! = (2\beta)^2,\,$ so check $\,\alpha \to 2\beta\,$ preserves sums and products.