make ggml_is_view_op public.

[foldl]

Motivation: ggml_is_view_op is a useful API.

Use case 1

It is used by test-backend-ops.cpp.

Use case 2

llama.cpp/src/llama.cpp

Lines 8178 to 8179 in 73e2ed3

	if (ggml_backend_supports_op(backend.get(), cur)) {
	ggml_backend_sched_set_tensor_backend(lctx.sched.get(), cur, backend.get());

Let's say cur is a view operation and its source is on another backend, then this would be problematic.
if (ggml_backend_supports_op(backend.get(), cur) && !ggml_is_view_op(cur->op)) seems better.

By the way, PR #10825 is useful for checking (visualizing) these scenarios.

[slaren]

This function, if made public, does not belong in ggml-backend.cpp. The obvious location would be ggml.c, but moving it there would prevent inlining this function in ggml-backend.cpp, which is not great since it is used very often in ggml_backend_sched.

[foldl]

@slaren, I think view_op is a concept defined by backends, so it is ok to be in ggml-backend.cpp. Then, the question is: Is ggml_is_view_op a good name. How about change it to ggml_backend_is_view_op?