$f1(n)=n^{0.999999}logn\\ f2(n)=1000000n\\ f3(n)=1.000001^{n}\\ f4(n)=n^{2}$
We know order of growth of$f4(n)$ an exponential function is higher than all other given function, so it is largest among all.
Since $f3(n)$ is exponential and $f4(n)$ is quadratic then $f4(n)$< $f3(n)$, as quadratic grows slower than exponential.
$f2(n)$ is linear while $f4(n)$ is quadratic so $f2(n)$<$f4(n)$
Now we have to check between $f1(n)$ and $f2(n)$,
$f1(n)=n^{0.999999}logn, \ f2(n)=1000000n\cong n(constant \ can \ be \ neglected)$
$f2(n)=n=n^{0.999999}n^{0.000001}$
Common part can be removed from both $f1(n) and f2(n)$
Now $f1(n)=logn$ and $f2(n)=n^{0.000001}$
It is clear that for large values of n $f1(n)$ grows slower than $f2(n)$, another way to check this is, we can substitute $n=2^{m}$ in both the function then
$f1(m)=m$ and $f2(m)=2^{0.000001m}$, hence $f1(n)<f2(n)$
Hence increasing order is $f1(n)<f2(n)<f4(n)<f3(n)$
