Minima Maxima: 9a³y=x(4a-x)³

I think something is wrong with your given. If you check for x = a, y ≠ 3a. Here is the checking of your answer key.

Set x = a and solve for y
$9a^2 y = a(4a + a)^3$

$9a^2 y = 125a^4$

$y = \dfrac{125a^2}{9}$   ←   not equal to 3a
 

Kindly check the given.
 

$9a^3 y = x(4a - x)^3$
 

Differentiate y in terms of x using product formula
$9a^3 y' = -3x(4a - x)^2 + (4a - x)^3$
 

Simplify (optional)
$9a^3 y' = (4a - x)^2 [-3x + (4a - x)]$

$9a^3 y' = (4a - x)^2 (4a - 4x)$

$9a^3 y' = 4(4a - x)^2 (a - x)$
 

Determine the 2^nd derivative
$9a^3 y'' = -4(4a - x)^2 - 8(4a - x)(a - x)$

$9a^3 y'' = -4(4a - x)[ (4a - x) + 2(a - x) ]$

$9a^3 y'' = -4(4a - x)(6a - 3x)$

$9a^3 y'' = -12(4a - x)(2a - x)$
 

Set y' to zero for maxima and/or minima
$0 = 4(4a - x)^2 (a - x)$
 

For (4a - x)² = 0:

For a - x = 0

Here is the graph of the curve with a = 1