Try this:

)cl all
> )cl completely
> f:=operator 'f
> y:=operator 'y
> f1:=subst(D(f(x,y(x)),x),D(y(x),x)=f(x,y(x)))
> f2:=subst(D(f1,x),D(y(x),x)=f(x,y(x)))
> f3:=subst(D(f2,x),D(y(x),x)=f(x,y(x))) -- here is where an indeterminate 
> variable first appears
> kxy:=eval(D(f(x,z),[x,z]),z=y(x))
>

should produce

f[,1,2](%A,y(x))
>

-Alasdair

On Monday, January 20, 2014 9:58:07 PM UTC+11, Ralf Hemmecke wrote:
>
> On 01/20/2014 09:31 AM, Alasdair wrote: 
> > My input: 
> > 
> > f:=operator 'f 
> >> y:=operator 'y 
> >> kxy:=eval(D(f(x,z),[x,z]),z=y(x))  -- <- thanks to Waldek for this idea 
> >> 
> > 
> > The output: 
> > 
> > f[,1,2](%A,y(x)) 
>
> In fact, I cannot reproduce your output. Can you give a complete session? 
>
> Ralf 
>
> (6) -> )clear completely 
>    All user variables and function definitions have been cleared. 
>    All )browse facility databases have been cleared. 
>    Internally cached functions and constructors have been cleared. 
>    )clear completely is finished. 
> (1) -> f:=operator 'f 
>
>    (1)  f 
>                                                           Type: 
> BasicOperator 
> (2) -> y:=operator 'y 
>
>    (2)  y 
>                                                           Type: 
> BasicOperator 
> (3) -> kxy:=eval(D(f(x,z),[x,z]),z=y(x)) 
>
>    (3)  f    (x,y(x)) 
>          ,1,2 
>                                                     Type: 
> Expression(Integer) 
>
>

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