The coefficient of $y$  In the expansion of ${\left(y^2+\frac{c}{y}\right)}^5$  is

Given expansion  ${\left(y^2+\frac{c}{y}\right)}^5$  is

 We have general term in the expansion ${\left(x+a\right)}^n$

$(\therefore T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$be the expansion of ${(x+a)}^n)$  

$T_{r+1}={}^5{C}_r{\left(y^2\right)}^{5-r}{\left(\frac{c}{y}\right)}^r$  

$T_{r+1}={}^5{C}_r{\left(y^2\right)}^{5-r}{\left(c\right)}^r{\left(y^{-1}\right)}^r$

$T_{r+1}={}^5{\text{C}}_r{y}^{10-3r}{\text{\hspace{0.17em}c}}^r\mathrm{.......................}\left(1\right)$

$y^{10-3r}$ compare with $y^1$  because of the coefficient of $y$

$\Rightarrow y^{10-3r}=y^1$

Which contains $y^1$  if $10-3r=1$

$\therefore r=3$  this substitute in Eqn (1)

$T_{3+1}={}^5{\text{C}}_3{y}^{10-3\times 3}{\text{\hspace{0.17em}c}}^3$

$T_4={}^5{\text{C}}_3{y}^1{\text{\hspace{0.17em}c}}^3$ $\begin{array}{c}({\therefore }^n{C}_r=\frac{n!}{(n-r)!r!})\\ \end{array}$

$T_4=\frac{5\times 4}{2\times 1}y{\text{\hspace{0.17em}c}}^3=y10{\text{\hspace{0.17em}c}}^3$  

The coefficient of $y$ is $10{\text{\hspace{0.17em}c}}^3$