Step 1: Understand what is asked.
We have $f(x)=\frac{x^2\log(\cos x)}{\log(1+x)}$ for $x\neq 0$ and $f(0)=0$. We must decide whether $f$ is continuous, and if so whether it is differentiable, at $x=0$.
Step 2: Find the leading behaviour of each piece near $0$.
Using series, $\log(1+x)=x-\frac{x^2}{2}+\cdots$, and $\cos x=1-\frac{x^2}{2}+\cdots$ so $\log(\cos x)=-\frac{x^2}{2}+\cdots$.
Step 3: Build the ratio to lowest order.
Then $f(x)\approx \frac{x^2\left(-\frac{x^2}{2}\right)}{x}=-\frac{x^3}{2}+\cdots$ for small $x$.
Step 4: Check continuity.
As $x\to 0$, $-\frac{x^3}{2}\to 0$, which equals $f(0)=0$. So $\lim_{x\to 0}f(x)=f(0)$ and $f$ is continuous at $0$.
Step 5: Check differentiability through the difference quotient.
The derivative at $0$ is $\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}\frac{f(x)}{x}=\lim_{x\to 0}\frac{-\frac{x^3}{2}}{x}=\lim_{x\to 0}\left(-\frac{x^2}{2}\right)=0$.
Step 6: Conclude.
The difference quotient has a finite limit, so $f'(0)$ exists. Hence $f$ is differentiable at $x=0$, which is the strongest of the listed options. \[ \boxed{\text{differentiable}} \]