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There are 2 different second order pade approximations of delay given in internet

What is the difference between these two approximation? Which one is the correct Pade approximation

The first one is not a Padé approximant but the ratio of two Taylor series.

Written as
$$e^{-x}=frac{e^{-x/2}}{e^{x/2}}=frac{1-frac{x}{2}+frac{x^2}{8}+Oleft(x^3right)}{1+frac{x}{2}+frac{x^2}{8}+Oleft(x^3right) }$$

This does not have anything to do with the second one which is the $[2,2]$ Padé approximant of $e^{-x}$

To show how is better the Padé approximant, let us compute
$$Phi_1=int_0^1 left(e^{-x}-frac{1-frac{x}{2}+frac{x^2}{8}}{1+frac{x}{2}+frac{x^2}{8}}right)^2,dxapprox 5.033times 10^{-5}$$ which, for sure, is better than
$$Phi_0=int_0^1 left(e^{-x}-left(1-x+frac{x^2}{2}-frac{x^3}{6}right)right)^2,dxapprox 1.375times 10^{-4}$$
but so worse than
$$Phi_2=int_0^1 left(e^{-x}-frac{1-frac{x}{2}+frac{x^2}{12}} {1+frac{x}{2}+frac{x^2}{12} }right)^2,dxapprox 3.127times 10^{-8}$$

Edit

Developed at a point $x=a$, for a given function $f(x)$, the $[m,n]$ Padé approximant write
$$p_{m,n}=frac{sum _{i=0}^m a_i (x-a)^i } {1+sum _{i=1}^n b_i (x-a)^i }$$ Then, let us write in a first step
$$left(1+sum _{i=1}^n b_i (x-a)^i right),f(x)=sum _{i=0}^m a_i (x-a)^i $$ and replace $f(x)$ by its corresponding Taylor expansion
$$f(x)= sum_{k=0} ^ {infty} frac {f^{(k)}(a)}{n!} (x-a)^{n}=sum _{k=0}^{infty} c_k (x-a)^k$$ which will be truncated to "some" order.

Now, expanding the summations and grouping for a given power of $(x-a)^j$, we have a series of linear equations in $a,b,c$ easy to solve for the $a$'s and $b$'s where the $c$'s are known. All of this is quite simple.

For illustration, let us do it for $f(x)=e^{-x}$ using the $[2,2]$ Padé approximant. We then have
$$(1+b_1 x+b_2 x^2)left(1-x+frac{x^2}{2}-frac{x^3}{6}+frac{x^4}{24}right)=a_0+a_1 x+a_2x^2$$

Expanding and grouping, this will give
$$(1-a_0)+ (-a_1+b_1-1)x+
left(-a_2-b_1+b_2+frac{1}{2}right)x^2+left(frac{b_1}{2}-b_2-frac{1}{6}right) x^3+frac{1}{24} (-4 b_1+12 b_2+1) x^4=0$$ Cancelling all coefficients, the gives the equations
$$1-a_0=0 qquad -a_1+b_1-1=0 qquad -a_2-b_1+b_2+frac{1}{2}=0qquad -4 b_1+12 b_2+1=0$$ the solutions of which being
$$a_0= 1qquad a_1=-frac 12 qquad a_2=frac 1 {12} qquad b_1=frac 12 qquad b_2= frac 1 {12}$$ which is your second expression.

Back to the two formulations given in your post, let us make the long divisions to get
$$frac{1-frac{x}{2}+frac{x^2}{8}}{1+frac{x}{2}+frac{x^2}{8} }=1-x+frac{x^2}{2}-frac{x^3}{8}+frac{x^5}{64}+Oleft(x^6right)=1-x+frac{x^2}{2}+Oleft(x^color{red}{3}right)$$ while
$$frac{1-frac{x}{2}+frac{x^2}{12}}{1+frac{x}{2}+frac{x^2}{12}}=1-x+frac{x^2}{2}-frac{x^3}{6}+frac{x^4}{24}-frac{x^5}{144}+Oleft(x^6right)=1-x+frac{x^2}{2}-frac{x^3}{6}+frac{x^4}{24}+Oleft(x^color{red}{5}right)$$ when compared to the Taylor expansion of $e^{-x}$.

I hope and wish that this makes things clearer for you. If not, just post.