Assumptions such as smoothness will not guarantee this. Consider $f(x,y)=\varphi((1+y^2)x)$ with a $\varphi\in C_0^{\infty}(\mathbb R)$ with $\varphi\ge 0$, $\varphi(0)=1$, $\int\varphi = 1/\pi$. Then $f\in C^{\infty}$, $$ \int dy\int dx f(x,y)= \frac{1}{\pi}\int \frac{dy}{1+y^2} =1 , $$ so $f$ is a density, but $f(0,y)=1\notin L^1$.

Assumptions such as smoothness will not guarantee this. Consider $f(x,y)=\varphi((1+y^2)x)$ with a $\varphi\in C^{\infty}(\mathbb R)$ with $\varphi> 0$, $\varphi(0)=1$, $\int\varphi = 1/\pi$. Then $f\in C^{\infty}(\mathbb R^2)$, $$ \int dy\int dx f(x,y)= \frac{1}{\pi}\int \frac{dy}{1+y^2} =1 , $$ so $f$ is a density, but $f(0,y)=1\notin L^1$.