$1/x^2$ is not uniformly continuous on $(0,1)$

The following is what I tried. I just start studying analysis. I am not sure if I am right. Please let me know if you see any mistake.

Prove $$f(x)=1/x^2$$ is not uniformly continuous on (0,1).

Definition : $$f(x)=1/x^2$$ is uniformly continuous on (0,1) if, given $$\epsilon>0$$, there exists a $$\delta>0$$ such that for all $$x,y\in(0,1)$$ with $$|x-y|<\delta$$, we have $$|\frac{1}{x^2}-\frac{1}{y^2}|<\epsilon$$.

So, my goal is to find a epsilon that contradicts the definition so that I prove $$f(x)$$ is not uniformly continuous.

Let $$x\in(0,1)$$.

Since $$0, we have $$1<\frac{1}{x}$$. So, $$1<\frac{1}{x^2} \Leftrightarrow 3<\frac{3}{x^2}$$.

And, Suppose $$y=\frac{x}{2}$$.

Then, $$|f(x)-f(y)|=|\frac{1}{x^2}-\frac{1}{y^2}|=|\frac{1}{x^2}-\frac{4}{x^2}|=|\frac{3}{x^2}|>3$$.

So, if we choose $$\epsilon =2$$ and $$y=x/2$$, $$|f(x)-f(y)|$$ is never less than $$\epsilon$$. Thus, it is not uniformly continuous.

(I feel like I should use $$\delta$$ somewhere , but not sure where to use it...please give me some advise.

You are correct. With regard your uncertainty regarding $$\delta$$...
You are saying for your chosen $$\epsilon$$, no $$\delta$$ will work. Suppose such a $$\delta$$ existed for your $$\epsilon$$. Then, if $$|x-y|=|x-\frac{x}{2}|=|\frac{x}{2}|<\delta$$, we have $$|f(x)-f(y)|>3>\epsilon$$. Most importantly, we need the existence of such an $$x$$ in our domain and for $$\frac{x}{2}$$ to also be in the domain. This works as $$(0,1)$$ is our domain and not say $$(1,\infty)$$.
In short, you implictly mention it by having $$\frac{x}{2}$$.
Assume that your function is uniformly continuous on $$(0,1)$$. Then for a given $$\varepsilon >0$$ it must exists a $$\delta=\delta(\varepsilon)>0$$ such that for every $$x,y \in (0,1)$$ such that $$|x-y|<\delta$$ it is $$|f(x)-f(y)|<\varepsilon$$. Now, let be $$x=2/n$$ and $$y=1/n$$ with $$n$$ positive integer greater than $$2$$. You will get $$\left| {x - y} \right| < \delta \Rightarrow \left| {\frac{1} {{x^2 }} - \frac{1} {{y^2 }}} \right| < \varepsilon$$ or $$\frac{{\left| {y - x} \right|\left| {y + x} \right|}} {{x^2 y^2 }} < \varepsilon$$ Since $$|x-y|=1/n$$, if $$n>1/\delta$$ you would have that for such values of $$n$$ it must be $$\frac{{\left| {\frac{1} {n} - \frac{2} {n}} \right|\left| {\frac{1} {n} + \frac{2} {n}} \right|}} {{\frac{1} {{n^2 }} \cdot \frac{4} {{n^2 }}}} < \varepsilon$$ or $$3n^2/4<\varepsilon$$ which is absurd.