You should use standard Binomial theorem exppansion here.$(a+b)^{n} = \binom{n}{0} a^{n} + \binom{n}{1} a^{n-1}b^{1} + \binom{n}{2} a^{n-2}b^{2}+...\binom{n}{n} b^{n}$.
Now in your case it become:
$$(2x-\frac{3}{x^2})^{4} = \binom{4}{0} (2x)^{4} - \binom{4}{1} (2x)^{3}(\frac{3}{x^2})^{1} + \binom{4}{2} (2x)^{2}(\frac{3}{x^2})^{2}-\binom{4}{3} (2x)^{1}(\frac{3}{x^2})^{3}+\binom{4}{4} (\frac{3}{x^2})^{4}$$
You should use standard Binomial theorem exppansion here.$(a+b)^{n} = \binom{n}{0} a^{n} + \binom{n}{1} a^{n-1}b^{1} + \binom{n}{2} a^{n-2}b^{2}+...\binom{n}{n} b^{n}$.
Now in your case it become: $$(2x-\frac{3}{x^2})^{4} = \binom{4}{0} (2x)^{4} - \binom{4}{1} (2x)^{3}(\frac{3}{x^2})^{1} + \binom{4}{2} (2x)^{2}(\frac{3}{x^2})^{2}-\binom{4}{3} (2x)^{1}(\frac{3}{x^2})^{3}+\binom{4}{4} (\frac{3}{x^2})^{4}$$
Now I shall let you conclude