Consider two maps $F:\mathbb{R}^{N}\rightarrow\mathbb{R}$ and $f:\mathbb{R}^{N}\rightarrow\mathbb{R}^{N}$. Notice that, is $f$ is the gradient of $F$, the following equation holds for every $x$:$$F(x)=F(0)+\int_{0}^{1}\sum_{n}x_{n}f_{n}(tx)\text{d}t.$$(to see this note that, if $f$ is the gradient of $F$, then $\text{d}F(tx)/\text{d}t=\sum_{n}x_{n}f_{n}(tx)$, so the above equation follows from the fundamental theorem of calculus).
Conversely, suppose that we know that the above Equation is satisfied for all $x$. Does this imply that $f$ is the gradient of $F$? Taking the derivative of both sides does not immediately yields the result:$$\frac{\text{d}F(x)}{\text{d}x_{n}}=\int_{0}^{1}f_{n}(tx)\text{d}t+\int_{0}^{1}\sum_{m}tx_{m}\frac{\text{d}f_{m}(tx)}{\text{d}x_{n}}\text{d}t.$$
For $N>1$ this is not true. Consider e.g. $$F\equiv 0$$ and $$f_1(x):=x_2,\quad f_2(x):=-x_1\quad \text{and}\quad f_n(x):=0\ \text{for }n>2.$$ Then, $$\sum_n x_nf_n(tx)=tx_1x_2-tx_2x_1=0$$ for all $x$ and $t$, thus your equation is satisfied. However, clearly $f$ isn't the gradient of $F$.
For $N=1$ on the other hand, the above construction doesn't work – and it actually works in one dimension: Note first that for all $t$ and $x$ we have $$\frac{\mathrm{d}}{\mathrm{d}x}\left(xf(tx)\right)=f(tx)+txf'(tx)=\frac{\mathrm{d}}{\mathrm{d}t}\left(tf(tx)\right).$$ Hence, differentiating your equation gives indeed $$F'(x)=\int_0^1\frac{\mathrm{d}}{\mathrm{d}x}\left(xf(tx)\right)\mathrm{d}t=\int_0^1\frac{\mathrm{d}}{\mathrm{d}t}\left(tf(tx)\right)\mathrm{d}t=f(x).$$