The likelihood function given the sample $\mathbf x=(x_1,\ldots,x_m,y_1,\ldots,y_n)$ is
$$L(\theta_1,\theta_2)=\theta_1^m\theta_2^n\exp\left(-\theta_1\sum_{i=1}^m x_i-\theta_2\sum_{i=1}^n y_i\right)\mathbf1_{x_1,\ldots,x_m,y_1,\ldots,y_n>0}\quad,\,\theta_1,\theta_2>0$$
Unrestricted MLE of $(\theta_1,\theta_2)$ is $$(\hat\theta_1,\hat\theta_2)=\left(\frac{1}{\bar x},\frac{1}{\bar y}\right)$$
Restricted MLE of $(\theta_1,\theta_2)$ when $\theta_1=\theta_2=\theta$ (say) is
$$\hat\theta=\frac{m+n}{m\bar x+n\bar y}$$
So the LR test statistic for testing $H_0$ is
$$\Lambda(\mathbf x)=\frac{\sup_{\theta_1=\theta_2}L(\theta_1,\theta_2)}{\sup_{\theta_1,\theta_2}L(\theta_1,\theta_2)}=\frac{L(\hat\theta,\hat\theta)}{L(\hat\theta_1,\hat\theta_2)}$$
Substituting the values of $\hat\theta_1,\hat\theta_2,\hat\theta$, the terms in the exponent of $e$ vanish, and I get
$$\Lambda(\mathbf x)=\bar x^m\bar y^n\left(\frac{m+n}{m\bar x+n\bar y}\right)^{m+n}$$
The above can be rewritten to get a 'nice' form:
\begin{align}
\Lambda(\mathbf x)&=\underbrace{\text{constant}}_{>0}\left(\frac{m\bar x}{m\bar x+n\bar y}\right)^m\left(\frac{n\bar y}{m\bar x+n\bar y}\right)^n
\\&=\text{constant}\cdot\,t^m(1-t)^n\qquad,\text{ where }t=\frac{m\bar x}{m\bar x+n\bar y}
\\&=g(t)\quad,\,\text{say}
\end{align}
Now you have to study the nature of the function $g$, keeping in mind that we reject $H_0$ when $\Lambda(\mathbf x)<c$ for some $c$ subject to a level restriction.
As for why this should be an $F$ test, note that
$$X_i\stackrel{\text{ i.i.d }}\sim\text{Exp with mean }1/\theta_1\implies 2\theta_1 X_i\stackrel{\text{ i.i.d }}\sim\text{Exp with mean }2\equiv \chi^2_2$$
So summing over independent chi-square variables, $$2m\theta_1\overline X\sim\chi^2_{2m}$$
Similarly, $$2n\theta_2\overline Y\sim\chi^2_{2n}$$
And since both samples are independent, our test statistic is
$$\frac{\theta_1\overline X}{\theta_2\overline Y}\sim F_{2m,2n}$$
