It's so easy !!

Let $u = \mathrm{f}(e^x)$ and $v = e^{\mathrm{f}(x)}$ .

Then $\frac{\mathrm{d}u}{\mathrm{d}x} = \mathrm{f}'(e^x)$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = \mathrm{f}'(x) \times e^{\mathrm{f}(x)}$ .

By the product rule, $\frac{\mathrm{d}y}{\mathrm{d}x} = v\frac{\mathrm{d}u}{\mathrm{d}x} + u\frac{\mathrm{d}v}{\mathrm{d}x}$ .

Substituting in the expressions from above, we get $\frac{\mathrm{d}y}{\mathrm{d}x} = e^{\mathrm{f}(x)} \times \mathrm{f}'(e^x) + \mathrm{f}(e^x) \times \mathrm{f}'(x) \times e^{\mathrm{f}(x)}$ .

From the information in the question, when $x = 0$ this reduces to $\frac{\mathrm{d}y}{\mathrm{d}x} = e^0 \times \mathrm{f}'(1) + \mathrm{f}(1) \times \mathrm{f}'(0) \times e^0$ , which simplifies further to $\frac{\mathrm{d}y}{\mathrm{d}x} = 1 \times 2 + 0 \times \mathrm{f}'(0) \times 1$ which is clearly equal to $\boxed{2}$ .