Yes, because
$$\text{Corr}(X,Y)\ne0 \Rightarrow \text{Cov}(X,Y)\ne0$$
$$\Rightarrow E(XY) - E(X)E(Y) \ne 0 $$
$$\Rightarrow \int \int xyf_{X,Y}(x,y)dxdy -\int xf_X(x) dx\int yf_Y(y)dy \ne 0$$
$$\Rightarrow \int \int xyf_{X,Y}(x,y)dxdy -\int \int xyf_X(x) f_Y(y)dxdy \ne 0$$
$$\Rightarrow \int \int xy \big[f_{X,Y}(x,y) -f_X(x) f_Y(y)\big]dxdy \ne 0$$
which would be impossible if $f_{X,Y}(x,y) -f_X(x) f_Y(y) =0,\;\; \forall \{x,y\}$. So
$$\text{Corr}(X,Y)\ne0 \Rightarrow \exists \{x,y\}:f_{X,Y}(x,y) \ne f_X(x) f_Y(y)$$
Question: what happens with random variables that have no densities?
Best Answer
Stock returns are a decent real-life example of what you're asking for. There's very close to zero correlation between today's and yesterday's S&P 500 return. However, there is clear dependence: squared returns are positively autocorrelated; periods of high volatility are clustered in time.
R code:
The timeseries of log returns on the S&P 500:
If returns were independent through time (and stationary), it would be very unlikely to see those patterns of clustered volatility, and you wouldn't see autocorrelation in squared log returns.