To find the Taylor series of $f(x) = 2x e^{e^{(3x)^2}}$ around $a = 0$, we need to compute the derivatives of $f(x)$ at $x = 0$ and use the Taylor series formula:

$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$

Let's proceed by evaluating the function and its derivatives at $x = 0$.

Step 1: Evaluate $f(0)$

$f(0) = 2 \cdot 0 \cdot e^{e^{(3 \cdot 0)^2}} = 0$

Since $f(0) = 0$, the constant term in the Taylor series is zero.

Step 2: Compute the First Derivative, $f'(x)$

Using the product rule for derivatives: $f(x) = 2x \cdot e^{e^{9x^2}}$ $f'(x) = 2 \cdot e^{e^{9x^2}} + 2x \cdot \frac{d}{dx}\left(e^{e^{9x^2}}\right)$

To differentiate $e^{e^{9x^2}}$ with respect to $x$, we use the chain rule: $\frac{d}{dx}\left(e^{e^{9x^2}}\right) = e^{e^{9x^2}} \cdot \frac{d}{dx}(e^{9x^2}) = e^{e^{9x^2}} \cdot e^{9x^2} \cdot (18x)$

Therefore, $f'(x) = 2 e^{e^{9x^2}} + 2x \cdot e^{e^{9x^2}} \cdot e^{9x^2} \cdot 18x = 2 e^{e^{9x^2}} + 36x^2 e^{e^{9x^2}} e^{9x^2}$

Now, evaluate $f'(0)$: $f'(0) = 2 e^{e^0} = 2e$

So, the coefficient of $x$ in the Taylor series is $2e$.

Step 3: Compute the Second Derivative, $f''(x)$

This calculation involves differentiating $f'(x)$ and becomes quite complex. However, for practical purposes, you might only need the first few terms.

After computing higher derivatives, you