If $$f(1) = 1, f'(1) = 3$$, then the derivative of $$f(f(f(x))) + (f(x))^{2}$$ at $$x = 1$$ is:

We have to find the derivative at $$x = 1$$ of the function

$$g(x)=f\!\left(f\!\left(f(x)\right)\right)+\left(f(x)\right)^{2}.$$

The given data are

$$f(1)=1 \qquad\text{and}\qquad f'(1)=3.$$

First we tackle the derivative of the composite term $$f\!\left(f\!\left(f(x)\right)\right).$$

By the chain rule, which states that if $$h(x)=f\!\bigl(u(x)\bigr)$$ then $$h'(x)=f'\!\bigl(u(x)\bigr)\,u'(x),$$ we proceed step by step through every layer of the composition.

Let us denote successively

$$u(x)=f(x),$$

$$v(x)=f(u(x))=f\!\left(f(x)\right),$$

so that

$$f\!\left(f\!\left(f(x)\right)\right)=f(v(x)).$$

Applying the chain rule twice in succession we write

$$\frac{d}{dx}\,f\!\left(f\!\left(f(x)\right)\right) =f'\!\bigl(v(x)\bigr)\cdot v'(x).$$

Now we still need $$v'(x).$$ Again by the chain rule,

$$v'(x)=\frac{d}{dx}\,f\!\left(f(x)\right)=f'\!\bigl(u(x)\bigr)\cdot u'(x).$$

Finally $$u'(x)=f'(x)$$ directly from the definition of $$u(x).$$ Collecting all three factors, the complete derivative of the triple composition is

$$\frac{d}{dx}\,f\!\left(f\!\left(f(x)\right)\right) =f'\!\bigl(v(x)\bigr)\;f'\!\bigl(u(x)\bigr)\;f'(x).$$

In compact form, using the original names, this is

$$\frac{d}{dx}\,f\!\left(f\!\left(f(x)\right)\right) =f'\!\Bigl(f\!\bigl(f(x)\bigr)\Bigr)\;f'\!\bigl(f(x)\bigr)\;f'(x).$$

Now we evaluate this at $$x=1.$$

First compute the necessary inner values:

$$f(1)=1\quad\text{(given)}.$$

Therefore $$f\!\bigl(f(1)\bigr)=f(1)=1$$ again.

Now substitute these into each derivative factor one by one:

$$f'\!\Bigl(f\!\bigl(f(1)\bigr)\Bigr) =f'(1)=3,$$

$$f'\!\bigl(f(1)\bigr)=f'(1)=3,$$

and directly $$f'(1)=3.$$

Hence

$$\left.\frac{d}{dx}\,f\!\left(f\!\left(f(x)\right)\right)\right|_{x=1} =3 \times 3 \times 3 = 27.$$

Next, we differentiate the square term $$\bigl(f(x)\bigr)^{2}.$$ Using the rule for a square, namely $$\displaystyle\frac{d}{dx}\bigl(u^{2}\bigr)=2u\,u',$$ with $$u=f(x),$$ we get

$$\frac{d}{dx}\left(f(x)\right)^{2}=2\,f(x)\,f'(x).$$

Evaluating at $$x=1$$ gives

$$2\,f(1)\,f'(1)=2 \times 1 \times 3 = 6.$$

Now we add the two derivative parts to obtain the derivative of the whole function $$g(x)$$ at $$x=1$$:

$$g'(1)=27+6=33.$$

Hence, the correct answer is Option D.