The number of pairs $(m,n)$ of coprime positive integers that divide $k$ is $d(k^2)$, where $d$ is the divisor counting function. I recently found somewhere that, if $k$ is a fixed integer.Then the number of ordered pairs of positive integers $(m,n)$ such that they are coprime and both of them divide $k$ is $d(k^2)$, where $d$ denotes "the number of divisor function."I tried but was unable to prove it.  I thought there might be some bijection between them but failed to prove it.
 A: Let the count of admissible pairs for $k$ be $F(k)$. Following lulu's hint in comments:


*

*For $k=p^n$ a prime power, admissible pairs can only be $(1,p^z)$ and $(p^z,1)$, $0\le z\le k$. There are $2k+1$ such pairs and $\tau(k^2)=\tau(p^{2n})=2k+1$.

*Let $F(a)=\tau(a^2),F(b)=\tau(b^2)$ and $\gcd(a,b)=1$. Let $(a_1,a_2)$ be an admissble pair for $a$, and similarly for $(b_1,b_2)$ and $b$. Then $(a_1b_1,a_2b_2)$ is an admissible pair for $ab$ since $a_1,a_2,b_1,b_2$ are pairwise coprime. Furthermore, any admissible pair for $ab$ can be uniquely decomposed into a pair for $a$ and a pair for $b$ by matching prime factors; for example, if $a=21$ and $b=10$, the pair $(14,15)$ which is admissible for $ab$ is decomposed into $(7,3)$ for $a$ and $(2,5)$ for $b$. Thus
$$F(ab)=F(a)F(b)=\tau(a^2)\tau(b^2)=\tau((ab)^2)$$
where we have used the multiplicative property of $\tau$.


This shows that $F(k)=\tau(k^2)$ for all positive $k$.
A: For each positive integer $K$, let $D(K)$ denote the set of all positive integers that divide $K$, and $S(K)$ the set of all pairs $(M,N)$ of relatively prime positive integers that divide $K$.  For a given positive integer $k$, we shall establish a bijection $f:D(k^2)\to S(k)$ along with its inverse $g:S(k)\to D(k^2)$. Let $$k=p_1^{r_1}p_2^{r_2}\cdots p_l^{r_l}\,,$$ where $p_1,p_2,\ldots,p_l$ are distinct prime natural numbers, and $r_1,r_2,\ldots,r_l$ are positive integers. 
Each $s\in D(k^2)$ is of the form $s=p_1^{t_1}p_2^{t_2}\cdots p_l^{t_l}$, where $t_i$ is an integer such that
$$0\leq t_i \leq 2r_i\text{ for every }i=1,2,\ldots,l\,.$$
Define
$$\mu(s):=\prod_{\substack{i\in\{1,2,\ldots,l\}\\ t_i\leq r_i}}\,p_i^{t_i}\text{ and }\nu(s):=\prod_{\substack{i\in\{1,2,\ldots,l\}\\ t_i> r_i}}\,p_i^{t_i-r_i}\,.$$
Note that $\mu(s)$ and $\nu(s)$ are divisors of $k$ and $\gcd\big(\mu(s),\nu(s)\big)=1$.  Therefore, if we set
$$f(s):=\big(\mu(s),\nu(s)\big)$$
for each $s\in D(k^2)$, then $f:D(k^2)\to S(k)$.  
Suppose now that $(m,n)\in S(k)$.  Write
$$n=p_1^{\beta_1}p_2^{\beta_2}\cdots p_l^{\beta_l}\,,$$
where $\beta_i$ is an integer such that
$$0\leq \beta_i\leq r_i\text{ for each }i=1,2,\ldots,l\,.$$
Define
$$g(s):=m\,\prod_{\substack{i\in\{1,2,\ldots,l\}\\\beta_i>0}}\,p_i^{r_i+\beta_i}\,.$$
Then, clearly, $g:S(k)\to D(k^2)$ is the inverse function of $f$.  Thus,
$$\sigma_0(k^2)=\big|D(k^2)\big|=\big|S(k)\big|\,.$$
(I prefer the notation $\sigma_0$, instead of $d$.)
Remark.  In general, for any positive integer $t$, $\sigma_0(k^t)$ is the number of tuples $(n_1,n_2,\ldots,n_t)$ of pairwise relatively prime positive integers such that $n_j$ is a divisor of $k$ for all $j=1,2,\ldots,t$.  My proof for the case $t=2$ can be extended easily to the general situation.  A more difficult question is: what is the number of tuples $(n_1,n_2,\ldots,n_t)$ of positive integers such that $n_j$ is a divisor of $k$ for all $j=1,2,\ldots,t$, and $\gcd(n_1,n_2,\ldots,n_t)=1$?
