Derivative of $h(x,t)=gleft(frac{x}{t^2}right)$

I think the op has improved his post , thus i could suggest this trivial solution :

$ h(x,t)=g(y)=g(x/t^2) $

$ frac{partial h(x,t) }{partial t} = g^{'}(y(t)) times y^{'}(t) $

$ y^{'}(t)= -2x/t^3 $

g is a function of y , and y is a function of t so it will be simple to continue the other steps ! the derivative of y is already given in the previous answer !

If i understood the situation well ,this may help :

Define $y= g_1(x,t) = x / t^2$.

Such that $h(x,t)=g(y)=g(g_1(x,t))$.

Then $frac{partial g_1 }{partial t}= frac{partial y }{partial t}=frac{-2times t times x }{t^4 }$,

with $frac{partial y}{partial t}=frac{partial g_1}{partial t}=-2 : frac{x}{t^3}$ where $x neq x(t)$.

We assume here that x is independant of t .

Then :

$$ frac{partial h(x,t) }{partial t}= frac{partial g }{partial y} times frac{partial g_1 }{ partial t } $$

And :

$$ g^{'}(y) = frac{partial g }{partial y} $$

$$frac{partial h}{partial x}=g'(g_1(x,t))cdot frac{partial g_1}{partial x}(x,t)=g'left(frac{x}{t^2}right)cdot frac{1}{t^2}$$ and $$frac{partial h}{partial t}=g'(g_1(x,t))cdot frac{partial g_1}{partial t}(x,t)=g'left(frac{x}{t^2}right)cdot frac{(-2x)}{t^3}$$

Edit: Note that when you wrote the formula for partial derivative of $f,$ you had $2$ dependent functions $x$ and $y,$ whereas your function is actually simpler in that you only have one dependent function $g_1.$ Drawing a tree diagram (for example, explained here) always helped me during undergrad while applying chain rule when dealing with partial derivatives.