Showing there exists a point for which gradient is zero

Question

I have the following homework question:

$f$ is a scalar function, differentiable everywhere. $h$ is a function defined as follows: $h(t,s) = tsf(ts,t-s)$. Show that there exists a point $(t,s)$ $\neq$ $(0,0)$ for which $\nabla h(t,s) = (0,0)$

My attempt:

$\nabla h(t,s) = ((s(f(ts,t-s)+tf_t(ts,t-s)), t(f(ts,t-s), sf_s(ts,t-s)))$

$\nabla h(t,s) = (0,0)$ so,

$(s(f(ts,t-s)+tf_t(ts,t-s)) = 0$

$s \neq 0 $ so,

$(f(ts,t-s)+tf_t(ts,t-s)) = 0$

similarly,

$(f(ts,t-s)+sf_s(ts,t-s)) = 0$

I substructed the two equations and got:

$tf_t(ts,t-s) = sf_s(ts,t-s)$

Here is where i am stuck. How can i prove there exists $(t,s)$ that satisfy this equiation?

Thank you for your help.

Answer 1

Your calculation is wrong. $h_t=sf(ts,t-s)+ts(sf_1(ts,t-s)+f_2(ts,t-s))$ and $h_s$ is similar. Here $f_1$ and $f_2$ is partial derivatives.

Generally, $f_t(x(t),y(t))=x'(t)f_1(x(t),y(t))+y'(t)f_2(x(t),y(t))$.