Just inequality

$\frac{\left(x+y\right)^{3} }{x^{2}y}=\frac{\left(x+y\right)^{2}\left(x+y\right) }{x^{2}y}$

$\frac{\left(x+y\right)^{2}}{x^{2}} \times \frac{\left(x+y\right)}{y}$

$\left(\frac{x+y}{x}\right)^{2} \times \left(\frac{x}{y} +1 \right)$

$\left(\frac{y}{x} +1\right)^{2} \times \left(\frac{x}{y} +1 \right)$

let $\frac{y}{x} =k$

$y=kx$

$\left(k+1\right)^{2} \times \left(\frac{1}{k} +1 \right)$ simplifying further we get $\frac{\left( k+1 \right)^{3}}{k^{2}}$

if we plot this as a function and analyse the rate of change and solve the inequalities of rate of change being<0 and >0 we get unique number satisfying the inequality and observe that for values less than this number it is negative and for others it is positive, therefore the critical point is a minimum value. $\frac{3 \left(k+1 \right)^{2}}{k^{2}} -\frac{2 \left(k+1 \right)}{k^{3}} =0$

$3=2\frac{k+1}{k}$

$3k=2k+2$

$k=2$

substituting this value to the function

$\frac{\left( 2+1 \right)^{3}}{2^{2}}$

$\frac{27}{4} = 6.75$

Let $y=vx$ . Then the given expression is reduced to $v^2+3v+3+\dfrac{1}{v}$ . This will be minimum when $v=\dfrac{1}{2}$ (since both $x, y$ and hence $v$ are positive). Hence the minimum value of the given expression is $\dfrac{27}{4}=\boxed {6.75}$

Let $m=(x+y)$

The formula is now

$m^3/(x^2)(m-x)$

For a given $m$ , we need to find the Highest value of the denominator $(x^2)(m-x)=mx^2-x^3$

This is a cubic equation passing through (0,0) and tending to -inf as $x$ tends to +inf.

By calculus the x-positive stationary point:

$2mx-3x^2=0$

$x=2m/3$

Substitute back into $m^3/(x^2)(m-x)$

All the algebra cancels out and the answer is $\boxed{27}$

Why does all the algebra cancel out? - Expand the formula in the question and it can be seen that every term is order-3 in $y$ and $x$ so $y$ and $x$ can vary in proportion without affecting the value.