Recall the binomial theorem:
[tex]\displaystyle (a + b)^n = \sum_{k=0}^n \binom nk a^{n-k} b^k[/tex]
where [tex]\binom nk = \frac{n!}{k!(n-k)!}[/tex] is the binomial coefficient.
Take a = x/3, b = -3, and n = 7. Then we get the x⁵ and x⁴ terms when 7 - k = 5 and 7 - k = 4, respectively; or when k = 2 and k = 3.
[tex]k=2 \implies \dbinom 72 \left(\dfrac x3\right)^{7-2} (-3)^2 = 21 \cdot \dfrac{x^5}{243} \cdot 9 = \dfrac79 x^5[/tex]
[tex]k=3 \implies \dbinom 73 \left(\dfrac x3\right)^{7-3} (-3)^3 = 35 \cdot \dfrac{x^4}{81} \cdot (-27) = -\dfrac{35}3 x^4[/tex]
Then when multiplying this expansion by x - 6, we get an x⁵ terms from the products
[tex]\dfrac79 x^5 \cdot (-6)[/tex]
and
[tex]-\dfrac{35}3 x^4 \cdot x[/tex]
so that the x⁵ term in the overall expansion of (x/3 - 3)⁷ (x - 6) has a coefficient of
[tex]\dfrac79\cdot(-6) + \left(-\dfrac{35}3\right) \cdot 1 = \boxed{-\dfrac{49}3}[/tex]
