Prove that a function is smooth if it is smooth in almost all directions

Question

So suppose we have a function $$f:\mathbb R^2\to \mathbb R$$ for which it is given that $$x\mapsto f(x,g(x))$$ is smooth (i.e., $$C^\infty$$) for all smooth functions $$g:\mathbb R\to\mathbb R$$. Can we prove that $$f$$ is smooth as well?

I don't know whether this statement is true and honestly I wouldn't be surprised either way.

What I've tried already

Fix a point $$(x_0,y_0)$$. Intuitively, by taking $$g(x) = \lambda x$$ with $$\lambda\in\mathbb R$$ we see that $$f$$ should be at least differentiable along all directions $$(1,\lambda)$$ at $$(x_0,y_0)$$. This follows for instance by considering the curve $$t\mapsto (x_0+t, y_0+\lambda t)$$. Thus the only direction that is non-trivial is the vertical direction $$(0,1)$$. If we can show that $$f$$ is also differentiable in that direction then I'm confident that it will be possible to show that $$f$$ is differentiable. But how can we show whether $$f$$ is differentiable along $$(0,1)$$? We cannot do it directly from the fact that $$f(x,g(x))$$ is smooth, but perhaps we can use a limiting argument, letting the slope of the curve $$(t,g(t))$$ tend to infinity?

When we know that $$f$$ is differentiable, it will probably be possible using an inductive argument to prove that $$f$$ is smooth (i.e., $$C^\infty$$).

Any help is appreciated.

EDIT. If found a closely related result, namely Boman's theorem, which says basically says that $$f$$ is smooth if and only if $$f\circ\gamma$$ is smooth for all smooth curves $$\gamma:\mathbb R\to\mathbb R^2$$. I feel like the statement of my question should probably be reducible to this theorem. The only difficulty is that we don't necessarily know if our $$f$$ is differentiable along vertical curves, but perhaps this follows in some way.

It need not even be continuous. Let $$f(x,y) = \frac{xy^2}{x^2+y^4}$$ for $$(x,y)\neq (0,0)$$ and $$f(0,0) = 0$$. This is discontinuous, since $$\lim_{t\rightarrow 0} f(t^2,t) = \frac{t^4}{t^4+t^4} = \frac{1}{2} \neq 0$$. Now, $$f(x,g(x))$$ is clearly smooth whenever $$g(0) \neq 0$$, so it remains to check the case $$g(0)=0$$. Then, $$g(x) = x\int_0^1g'(xt)dt =: xh(x)$$, and $$h$$ is clearly smooth. Now for $$x\neq 0$$, $$f(x,g(x)) = \frac{xg(x)^2}{x^2+g(x)^4} = \frac{x^3h(x)}{x^2(1+x^2h(x)^2)} = x\frac{h(x)}{1+x^2h(x)^2},$$ which smoothly extends to $$f(0,g(0)) = 0$$, since $$1+x^2h(x)^2\geq1$$.